I may have to throw this over to overflow, but I figured I would try here, first. This is a question in continuum theory, but it reduces to a combinatorial/graph-theoretical question. I am highly confident that this conjecture is both true and already known, so a reference would also be fine. It might be in Menger's Kurventheorie, but I can't get a pdf nor read German.
We define the simple $n$-od to be any union of $n$ (non-degenerate) paths $A_1, \dots, A_n$ in some space $X$ that are pairwise-disjoint except at a single, common point $p$ (in other words, an $n$-pointed asterisk). Shown below is a simple $3$-od in the plane.
We define the order of a point $p \in X$ for $X$ a topological space to be the infimum of cardinals $\alpha$ (though here we just care about finite vs. infinite) such that for every neighborhood (equivalently, open neighborhood) $V$ of $p$ there is a neighborhood (equiv. open neighborhood) $U \subset V$ of $p$ such that $\partial(U)$ has at most $\alpha$ points. For example, the $3$-od as drawn above has order $1$ at the end points, order $2$ at points along edges and order $3$ at the vertex point $p$.
That the topology of the simple $n$-ods agrees with the picture given above is largely contained in this theorem: The order of the vertex point of any simple $n$-od is $n$. The topology of a simple $n$-od always agrees with the 'standard one' in the plane, so there are no complications here.
A Peano Continuum is a compact, connected, locally connected metric space. There is a theorem by Karl Menger called the $n$-Beisatz Theorem:
If $X$ is a Peano Continuum and $p \in X$ such that its order is at least $n$, then $X$ contains a simple $n$-od centered at $p$.
Thus any point of infinite order in a Peano Continuum is the vertex of simple $n$-ods for all $n \in \mathbb{N}$.
The space on the left in the figure above is called the Hairy Point (i.e. a 'harmonic fan' the lengths of whose edges converges to $0$), while the space on the right is called the Null Comb. It's known that every Peano Continuum which isn't a (topological, finite) graph contains a Hairy Point or Null Comb. But I am also pretty sure the following, which is what I want to prove, is true:
If $p \in X$ has infinite order then $X$ contains a copy of the Hairy Point centered at $p$.
So assuming $p$ has infinite order, then for every $n$ there is a simple $n$-od $T_n$ contained in $X$ centered at $p$. The way that I have been trying to prove the theorem so far is to construct a sequence $S_k$ of simple $k$-ods centered at $p$ so that $S_{k+1} = S_k$ outside of a small neighborhood of $p$. Then I can take a limit and prove the theorem in a routine manner. But constructing $S_{k+1}$ given $S_k$ hasn't worked out so well. The complication that arises is as follows.
We call the 'main' $n$ arcs in the simple $n$-od to be its chords. If $A, B$ are two arcs with end point $p$, we say that $A$ is dense along B at $p$ if $p$ is an accumulation point of $A \cap B$, as shown below.
(Easy Case): So if $T_k$ is a simple $k$-od centered at $p$ and there is some $T_n$ with $n > k$ such that it has a chord $A$ which is not dense in any of the chords $A_1, \dots, A_k$ of $T_k$, then after taking a sub-arc of $A$ we can assume that $A$ doesn't intersect any of the $A_j$ at all and thus we obtain a simple $(k+1)$-od $A \cup A_1 \cup \cdots \cup A_k$.
(Hard Case): Thus we may as well assume that for every $n$ with $n > k$ we have that every chord of $T_n$ is dense at $p$ along some chord of $T_k$ (note that it may be dense along several chords of $T_k$, e.g. by spiraling). Again we can assume that each chord of $T_n$ only intersects those $A_j$ that it's dense along. By setting $n = (k+1) \cdot 2^k + 1$, then noting that $2^k$ is the number of subsets of chords of $T_k$ by the pigeonhole principle there is some subset $A_{m_1}, \dots, A_{m_n}$ of chords of $T_k$ such that there are at least $k+1$ chords , and thus $m_n + 1$ chords (shown in red below), that are dense along all of $A_{m_1}, \dots, A_{m_n}$. By substituting the number of ordered subsets of $\lbrace 1, 2, \dots, k \rbrace$ we may assume that they first hit the chords $A_{m_1}, \dots, A_{m_n}$ in the same order (though there may be various manner of backtracking). Let $R$ be as shown.
Finally, at this point the problem really is just combinatorial. I have been trying various methods of removing sub-arcs of those relevant chords to obtain an $(m_n + 1)$-od but failing. Note that the picture is slightly simplified, since for each $A_{m_j}$ we have no control over the order in which the chords of $R$ intersect it. So I am wondering if the finite picture just isn't sufficient and it's some infinite combinatorial situation.
Anyone have any ideas? Ultimately, I am trying to generalize the following procedure to arbitrarily many blue chords.
If $X$ is a Peano continuum and ord$_X(p) \geq \aleph_0$ then $X$ contains a copy of the hairy point.
Proof: By the $n$-Beisatz Theorem there are simple $n$-ods $T_n \subset X$ centered at $p$. Fix a $k$; we claim that there is a simple $(k+1)$-od $S \subset X$ centered at $p$ such that $S = T_n$ outside the $\frac{1}{k+1}$-neighborhood of $p$.
(Case 1): First assume there is an $n$ so that some chord $B$ of $T_n$ is not dense at $p$ along any chord in $T_k$, whose chords we denote by $A_1, A_2, \dots, A_k$. Then by taking a (non-degenerate) sub-arc $B'$ of $B$ if necessary, we may assume that $B'$ is contained in the $\frac{1}{k+1}$-neighborhood of $p$ and $B' \cap A_j = \lbrace p \rbrace$ for all $j$. Thus $B' \cup T_k = S$ is a simple $(k+1)$-od centered at $p$, $T_k \subset S$ and $S \setminus T_k$ is an arc contained in the $\frac{1}{k+1}$-neighborhood of $p$, as desired.
(Case 2): Now assume that for every $T_n$ with $n > k$ we have that every chord of $T_n$ is dense along at least one of $A_1, \dots, A_k$ at $p$ (such a chord may be dense along more than one $A_j$ at $p$, for example by spiraling). As before, for each $T_n$ we may replace it by a simple $n$-od $T_n'$ so that every arc $B$ of $T_n'$ is contained in the $\frac{1}{k+1}$-neighborhood of $p$ and only intersects those $A_j$ along which it is dense at $p$.
Noting that $2^k$ is the number of subsets of $\lbrace 1, 2, \dots, k \rbrace$, by the pigeonhole principle if $n = (k+1) \cdot 2^k$ then for at least one subset $\lbrace m_1, \dots, m_\ell \rbrace$ of $\lbrace 1, 2, \dots, k \rbrace$ the simple $n$-od $T_n'$ contains some $k + 1$ chords $C_1, \dots, C_{k+1}$ each dense along all $A_{m_j}$ at $p$. In particular, there are at least $\ell + 1$ of them $C_1, \dots, C_{\ell + 1}$. Without loss of generality assume $m_j = j$ for $1 \leq j \leq \ell$, and let $R = C_1 \cup \cdots \cup C_{\ell + 1}$ be the simple $(l+1)$-od centered at $p$. Recall that its chords only intersect the chords of $T_k$ along which it is dense at $p$, i.e. $A_1, \dots, A_\ell$.
Assume each $A_j$ is ordered so that its initial point is $p$. Fix an $A_j$. Among the first $\ell$ chords $C_1, \dots, C_\ell$ of $R$, let $i$ be the index such that $A_j \cap C_i$ is the \textit{last} point of $(C_1 \cup \cdots \cup C_\ell) \cap A_j$ with respect to the order on $A_j$, and denote it by $c(A_j)$. This index exists since the $C_i$'s are closed and disjoint. Consider the set $\lbrace c(A_1), \dots, c(A_\ell) \rbrace$.
Assume there is an index $h$, for $1 \leq h \leq \ell$, such that $h \notin \lbrace c(A_1), \dots, c(A_\ell) \rbrace$. Then some collection of distinct chords $A_{j_1}, \dots, A_{j_h}$ of $T_k$ are last intersected by the same $C_i$ amongst $C_1, \dots, C_\ell$. Assume it is $i$; let $f_i: I \rightarrow C_i$ be a homeomorphism with $f_i(0) = p$, and consider the sub-arcs $f_i([0,t])$. If $C_i$ \textit{last} intersects $A_{j_1}, \dots, A_{j_h}$ in that order, by closedness of $C_i$ there is a maximal $t_0 \in (0,1)$ such that $f_i(t_0) \in A_{j_1}$.
By continuity of $f_i$, for $s > 0$ there are only finitely many times the set $f_i([s,1])$ changes which $A_{j_r}$ it last intersects (lest those times have an accumulation point, impossible since the $A_j$'s are mutually separated outside any neighborhood of $p$). If all of $A_{j_1}, \dots, A_{j_h}$ are still last intersected by $f_i([0,t_0])$ (as opposed to some $C_j$ with $j \neq i$), then since $C_h$ is dense along each $A_{j_m}$ there is an $x \in C_h \cap A_{j_m}$ and a $y \in C_i \cap A_{j_m}$ such that $y < x$ with respect to the ordering on $A_{j_m}$. Letting $s = f_i^{-1}(x) > 0$ we have, for any $j_m$, that after a finite number of excisions of terminal sub-arcs from $C_i$ we obtain $c(A_{j_m}) \neq i$. Thus after finitely many such excisions we may assume that $c(A_{j_m}) = i$ for precisely one $m$.
Replacing $C_i$ as above we obtain a new simple $(\ell+1)$-od $R' \subset R$ so that $i$ appears in $\lbrace c_{R'}(A_1), \dots, c_{R'}(A_\ell) \rbrace$ precisely once. Note that this replacement does not decrease the number of distinct elements of $\lbrace c(A_1), \dots, c(A_\ell) \rbrace$. If still $h \notin \lbrace c(A_1), \dots, c(A_\ell) \rbrace$ we may repeat this same procedure on some other $C_j$'s from $R'$, obtaining an $R''$ and proceeding inductively. Since for any $j$ there are only finitely many times $C_j$ changes which $A_i$ it last intersects, after finitely many iterations some $A_i$ will be last intersected by $C_h$. Thus after a finite number of steps we obtain a simple $(\ell+1)$-od $W$ satisfying $\lbrace c(A_1), \dots, c(A_\ell) \rbrace = \lbrace 1, 2, \dots, \ell \rbrace$. After taking a sub-arc, assume $C_{\ell+1}$ doesn't intersect any $A_{j_m}$ before one of the others does.
Order the first $\ell$ chords of $W$, which we also call $C_1, \dots, C_\ell$, so that they're in bijection with the chords $A_1, \dots, A_\ell$ such that $C_j$ is the chord of $W$ that last intersects $A_j$. For each pair $(A_j, C_j)$ let $S_j$ be the arc from $p$ to the other end point of $A_j$ that is the concatenation of $C_j$ up to the last point of $A_j \cap C_j$, followed by the remaining terminal sub-arc of $A_j$ as shown above. Let $S$ be the union of $C_{\ell+1}$, $S_j$ for $1 \leq j \leq \ell$ and $A_i$ for $i > \ell$.
We show that $S$ is a simple $(k+1)$-od centered at $p$, which will prove the claim. By construction it is the union of $k+1$ arcs each having $p$ as an end point; we must show this is the only point where any pair of intersect. If $i > \ell$ then $A_j$ doesn't intersect any other (alleged) chord of $S$ except at $p$ by construction of $T_n'$ and certainly no other $A_j$ does, since $T_k$ is a simple $k$-od. Since $C_{\ell+1}$ doesn't intersect any $A_i$ for $i > \ell$ except at $p$ by construction of $T_n'$, and doesn't intersect any $S_j$ by construction of $W$, it only remains to show that $S_i \cap S_j = \lbrace p \rbrace$ if $i \neq j$.
Writing $S_h = D_h \oplus B_h$, where $D_h$ is the portion of the chord $C_h$ and $B_h$ is the portion of the chord of $A_h$ comprising $S_h$, for $x \in S_i \cap S_j$, if $x \in D_i \cap D_j$ then $x = p$ since $W$ is a simple $(\ell+1)$-od, and since $B_i \cap B_j = \varnothing$ we may assume $x \in D_i \cap B_j$, without loss of generality. But then $C_i \cap A_j$ has a point of intersection after $C_j \cap A_j$, impossible by construction of $W$. Thus $S_i \cap S_j = \lbrace p \rbrace$ for $i \neq j$, proving the claim.
Returning to the main problem, by taking sub-arcs of the chords of $T_n'$ if necessary, assume $T_n'$ doesn't contain any end points (besides $p$) of the chords of $T_k$ (since they are bounded away from $p$). Thus we may assume each $S_j$ contains a non-trivial sub-arc of $A_j$ $(*)$. Similarly, we may assume that any $A_i$ is only intersected by a $C_j$ at a time (with respect to the order on $A_i$) before it first intersects the complement of the $\frac{1}{k+1}$-neighborhood $B$ of $p$ ($\dagger$) (the left figure is permitted; the right is not).
Thus there is a sequence of simple $n$-ods $T_n$ centered at $p$ such that, by ($\dagger$), $T_{n+1} = T_n$ outside the $\frac{1}{n+1}$-neighborhood of $p$ - we showed more, namely that by $(*)$ we may assume all but one chord of $T_{n+1}$ is a concatenation of a non-trivial arc and a non-trivial terminal sub-arc $B_i^{(n)}$ of a unique chord of $T_n$ (in Case 1 we simply subdivide each chord of $T_n$ into two non-trivial arcs). By taking sub-arcs if necessary, we may assume that $T_{n+2} \cap B_i = \varnothing$ for all $i$ such that $1 \leq i \leq n (\dagger \dagger)$.
We claim $\lim_n T_n = G$ is a hairy point centered at $p$. As $T_{n+1} = T_n$ outside the $\frac{1}{n+1}$-neighborhood of $p$, this limit exists since it's eventually constant outside every neighborhood of $p$, and $p \in T_n$ for all $n$. Thus $G$ is a continuum containing $p$. Setting notation, let $H$ be the hairy point with 'chords' $H_1, H_2, \dots$ where $H_n$ is the segment of length $\frac{1}{n}$ emanating from $p$. If $h_j$ is the linear homeomorphism of $I$ with $H_j$ with $h_j(0) = (0,0)$ then by the dyadics $2^{-n}$ in $H_j$ we mean the images under $h_j$ of the sequence $2^{-n}$ in $I$.
Let $f_0: T_2 \rightarrow H$ map $T_2$ homeomorphically to $H_1 \cup H_2$ with $f_0(p) = (0,0)$ and $f_0(A_j) = H_j$ for its chords $A_1, A_2$. By assumption, two chords $C_1, C_2$ of $T_3$ terminate in non-trivial sub-arcs $B_1^{(2)}, B_2^{(2)}$ of the chords $A_1, A_2$ of $T_2$ respectively.
Define $f_1: T_3 \rightarrow H$ so that $f_1(C_j) = H_j$ for $j = 1, 2, 3$, while $f_1(B_j) = [2^{-1}, 1]$ in $H_j$ as per the notation defined above for $j = 1, 2$. We may inductively define $f_n: T_{n+2} \rightarrow H$ so that it sends its chord $C_{n+2}$ homeomorphically onto $H_{n+2}$ with $f_n(p) = (0,0)$ and sends the corresponding $B_j^{(n+1)}$ of $T_{n+1}$ to $[2^{-n},1]$ in $H_j$. By $(\dagger \dagger)$ we have that $B_j^{(n)} \subset B_j^{(n+1)}$, since $T_{n+2}$ doesn't intersect $B_j^{(n)}$ for any $j$ with $1 \leq j \leq n$. Thus we may assume $f_n \equiv f_k$ on $B_j^{(k)}$ for all $j$ and for all $n \geq k + 2$.
If $x \in G = \lim_n T_n$ with $x \neq p$ then $x$ is contained in the (closed) complement $E$ of some $\frac{1}{n}$-neighborhood of $p$. Since $T_k \cap E$ is a sequence of closed sets that is eventually constant, $G \cap E = T_N \cap E$ for some $N$, which thus contains $x$. Thus $x$ is contained in $B_j^{(N+2)}$ for some $j$, on which the sequence $f_n$ is eventually constant, so $\lim_n f_n(x)$ is well-defined. If $x = p$ then since $f_n (p) \equiv 0$ we have $\lim_n f(p) = 0$, so $\lim_n f_n = f: G \rightarrow H$ is a well-defined function on $G$.
We wish to show that $f$ is a homeomorphism. If $x \in G$ with $x \neq p$ then as before $x \in B_j^{(N+2)}$ for some $j$ and for some $N$ sufficiently large; since the $f_n$'s are eventually constant, each a homeomorphism on $B_j^{(N+2)}$, we have that $f = \lim_n f_n$ is continuous at $x$. If $x_k \rightarrow p$ with $x_k \in G$ then eventually $x_k$ is contained in the $\frac{1}{n+2}$-neighborhood of $p$ and therefore mapped to $[0, 2^{-n}]$ in some $H_{j(k)}$ by definition of $f_n$. Thus $f$ is continuous at $p$, i.e. $f$ is continuous.
By construction, $f(G)$ contains the end point of $H_j$ for every $j$; since it also contains $p$ and is the continuous image of a connected set, $f(G)$ contains $H_j$ for all $j$. Since $H = H_1 \cup H_2 \cup \cdots$, we have that $f$ is surjective. If $x, y \in G$ are two distinct points other than $p$, then for $N$ sufficiently large we have $x \in B_{j(x)}^{(N+2)}$ and $y \in B_{j(y)}^{(N+2)}$. Since $f(B_{j(x)}^{(N+2)}) \cap f(B_{j(y)}^{(N+2)}) = \varnothing$ if $j(x) \neq j(y)$ and $f$ is a homeomorphism on any given $B_{j(x)}^{(N+2)}$ we have that $f(x) \neq f(y)$. Similarly, since $(0,0) \notin f(B_{j(x)}^{(N+2)})$ if $x = p$ and $ y \neq p$ we have $f(x) \neq f(y)$.
Thus $f$ is a continuous bijection of the compact $G$ and the Hausdorff $H$, and is therefore a homeomorphism.