Suppose X is a compact and connected metric space.
Then X is totally bounded and has a $\frac{\delta}{2}$-net, $\{x_1, \ldots, x_k\}$. I want to show, by induction, that we may arrange $\{x_1, \ldots, x_k\}$ (re-labelling as necessary) such that $d(x_i, x_{i+1}) \lt \delta, \forall i = 1, \ldots, k-1$.
Base Case: Suppose $k=2$. Then X = $B_{\frac{\delta}{2}}(x_1) \cup B_{\frac{\delta}{2}}(x_2)$. Since X is connected, we must have: $$B_{\frac{\delta}{2}}(x_1) \cap B_{\frac{\delta}{2}}(x_2) \neq \emptyset$$ , as otherwise X is not connected. Pick $y \in B_{\frac{\delta}{2}}(x_1) \cap B_{\frac{\delta}{2}}(x_2)$. Then, $d(y,x_1),d(y,x_2) \lt \frac{\delta}{2}$. So, $d(x_1,x_2) \lt \delta$.
Induction Hypothesis: Suppose that for $k \ge 2$, we may order a $\frac{\delta}{2}$-net $\{x_1, \ldots, x_k\}$ of X such that $d(x_i, x_{i+1}) < \delta, \forall i = 1, \ldots, k-1$.
Induction: Consider a $\frac{\delta}{2}$-net $\{x_1, \ldots, x_{k+1}\}$ of X. This is where I'm stuck. My approach thus far has been: By hypothesis we can arrange $\{x_1, \ldots, x_{k}\} \subset \frac{\delta}{2}$-net $\{x_1, \ldots, x_{k+1}\}$ such that $d(x_i, x_{i+1}) < \delta, \forall i = 1, \ldots, k-1$ (doubting that this part is correct). Then set $U = \cup_{i=1}^kB_{\frac{\delta}{2}}(x_i)$ and $V = B_{\frac{\delta}{2}}(x_{k+1})$. Then we must have $U \cap V \neq \emptyset$, as otherwise X is not connected. Choose $y \in U \cap V$. Then $d(y, x_{k+1}) \lt \frac{\delta}{2}$ and $\exists j \in \{1, \ldots, k\}$ such that $d(y, x_j) \lt \frac{\delta}{2}$. So, $d(x_{k+1}, x_j) \lt \delta$. From here I don't know where to go, as this does not directly show that we can arrange the whole set so that $d(x_i, x_{i+1}) < \delta, \forall i = 1, \ldots, k$.
Any help is really appreciated!
Let $\mathcal{N} = \{ x_1, ..., x_k \}$ be a $\frac{\delta}{2}$-net in a connected metric space $X$. That is, using the notation $U(x) = B_{\frac{\delta}{2}}(x)$ we have $\bigcup_{i=1}^k U(x_i) = X$. Such a finite set $\mathcal{N}$ exists for every $\delta > 0$ provided $X$ is totally bounded.
A $\delta$-sequence in $X$ is an $r$-tuple $(y_1,...,y_r)$, $y_i \in X$, such that $d(y_i,y_{i+1}) < \delta$ for all $i$.
A covering sequence for $\mathcal{N}$ is a $\delta$-sequence $(y_1,...,y_r)$ in $X$, such that $\{ y_1, ...,y_r\} = \mathcal{N}$. We have to distinguish the $r$-tupel $(y_1,...,y_r)$ from the set $\{ y_1, ..., y_r \} = \mathcal{N}$.
In the strict sense of the definition only sets can be $\frac{\delta}{2}$-nets, but for practical purposes there is no obstacle to use tuples $(y_1,...,y_r)$ instead of sets because $\bigcup_{i=1}^r U(y_i)$ is well-defined. In fact, we may regard a covering sequence $(y_1,...,y_r)$ for $\mathcal{N}$ as a $\frac{\delta}{2}$-net with repetitions.
Does there exist a covering sequence for $\mathcal{N}$ and what can be said about its length $r$? In particular, does there exist a covering sequence for $\mathcal{N}$ with length $k$, i.e can the $x_i$ be reindexed to obtain a covering sequence?
Let $k$ be fixed. We prove by induction that the following statement $A(n)$ is true for all $n$:
If $n \le k$, there exists a $\delta$-sequence $y_1,...,y_r$ of length $r = 1$ when $n = 1$, $r= 2$ when $n=2$ and $r = 2n - 3$ when $n \ge 3$ such that $\{ y_1, ...,y_r\}$ contains exactly $n$ elements of $\mathcal{N}$.
Then $A(k)$ tells us that there exists a covering sequence for $\mathcal{N}$. For $k\le 3$ we see that the $x_i$ can be reindexed to obtain a covering sequence, for $k > 3$ we can only say that there exist covering sequences with length $\le 2k-3$.
We come to the proof.
$A(1)$: Trivial.
$A(2)$ for $2 \le k$: We have $U(x_1) \cap \bigcup_{i=2}^k U(x_i) \ne \emptyset$ by connectedness of $X$. Then $U(x_1) \cap U(x_i) \ne \emptyset$ for some $i \ge 2$ and $(x_1,x_i)$ is the desired sequence (which has length $2$).
$A(3)$ for $3 \le k$: Take the sequence from $A(2)$ where we assume w.l.o.g. that $i = 2$. $U(x_1) \cup U(x_2)$ and $\bigcup_{i=3}^k U(x_i)$ cannot be disjoint by connectedness. Therefore $(U(x_1) \cup U(x_2)) \cap U(x_i) \ne \emptyset$ for some $i \ge 3$. If $U(x_1) \cap U(x_i) \ne \emptyset$, then $(x_i,x_1,x_2)$ is the desired sequence, if $U(x_2) \cap U(x_i) \ne \emptyset$, then $(x_1,x_2,x_i)$. It has length $3 = 2 \cdot 3 -3 $.
Assuming that $A(n)$ is true for $n \ge 3$, we prove $A(n+1)$ for $n+1 \le k$:
Take the sequence $(y_1,...,y_r)$ from $A(n)$. W.l.o.g. assume that $\{ y_1, ...,y_r\} = \{ x_1, ...,x_n \}$. $\bigcup_{i=1}^n U(x_i)$ and $\bigcup_{i=n+1}^k U(x_i)$ cannot be disjoint. We therefore find $s \in \{ 1, ...,n \}$ and $t \in \{ n+1, ...,k \}$ such that $U(x_s) \cap U(x_t) \ne \emptyset$. Choose $j \in \{ 1, ...,r \}$ such that $y_j = x_s$. Then $(y_1,...,y_j,x_t, y_j,...,y_r)$ is a sequence as required and its length is $r+2 = 2n - 3 + 2 = 2(n+1) - 3$.
For $k > 3$, can we find a covering sequence for $\mathcal{N}$ with length $< 2k-3$, in the best case with length $k$? In many cases yes, but in general $2k-3$ is best possible. Here is an example.
Let $e_1,...,e_{k-1}$ denote the standard basis of $\mathbb{R}^{k-1}$ and $L_i$ the line segments connecting $0$ and $e_i$. Define $X = \bigcup_{i=1}^{k-1} L_i$ with metric inherited from $\mathbb{R}^{k-1}$. If $1 < \delta < \sqrt{2}$, then $\mathcal{N} = \{ e_1, ..., e_{k-1}, 0\}$ is a $\frac{\delta}{2}$-net in $X$ with $k$ elements. We have $d(0,e_i) = 1, d(e_i,e_j) = \sqrt{2}$ for $i \ne j$. Let us try to find th shortest possible covering sequence for $\mathcal{N}$. If we start with $y_1 = 0$, then $y_2 = x_i$ for some $i$. All other $x_j$ are out of reach so that we must take $y_3 = 0$. Continuing in that way we see that we need $1 + 2(k-2) + 1 = 2k - 2$ points $y_i$ to cover $\mathcal{N}$ and that a smaller number is not sufficient. We can improve this if we start with $y_1 = x_i$. Then we save one point and get a covering sequence of length $2k-3$.
We can also give a graph theoretical interpretation of our results. For a $\frac{\delta}{2}$-net $\mathcal{N} = \{ x_1, ..., x_k \}$ define a graph $G(\mathcal{N})$ as follows: Its vertices are the points $x_i$ and its edges are all sets $\{ x_i,x_j \}$ such that $U(x_i) \cap U(x_j) \ne \emptyset$. $G(\mathcal{N})$ has a loop at each $x_i$. A walk in $G(\mathcal{N})$ is a finite sequence of vertices $(v_1, ..., v_r)$ of $G(\mathcal{N})$ such that each $\{ v_i,v_{i+1} \}$ is an an edge of $G(\mathcal{N})$.
There is 1-1-correspondence between $\delta$-sequences in $\mathcal{N} \subset X$ and walks in $G(\mathcal{N})$. Now we ask whether there exists a walk passing through all vertices. This holds true if and only if $G(\mathcal{N})$ is connected which means that for any two vertices $x_i , x_j$ there exists a walk from $x_i$ to $x_j$ (i.e. with $v_1 = x_i$ and $v_r = x_j$). It is easy to see that if $X$ is connected, then $G(\mathcal{N})$ is connected: If $G(\mathcal{N})$ would not be connected, then there would exist $x_i,x_j$ with no walk from $x_i$ to $x_j$. Let $A = \{ k \mid \text{There exists a walk from } x_i \text{ to } x_k\}$. Then $U = \bigcup_{k \in A} U(x_k)$ and $V = \bigcup_{k \notin A} U(x_k)$ are non-empty, open and disjoint which is imposible.