Consider a finite connected topological graph $G$ embedded on $\mathbb{R}^2$ with standard topology. Here vertices are points, and each edge is a Jordan arc between vertices; an embedding $e: [0, 1] \to \mathbb{R}^2$ such that $e(0)$ and $e(1)$ are vertices (we are excluding multi-edges and loops for simplicity) . The images of edges do not intersect except perhaps at vertices. An image of an edge does not intersect any other vertex than $e(0)$ and $e(1)$. Let a spike be a set of the form $s_v = \{\alpha v : 0 \leq \alpha < 1)\}$, where $v \in \mathbb{R}^2$. Note that $v = 0$ also provides a spike which is just the set $\{0\}$. Let a star be a set of the form $\cup_{i = 1}^n s_{v_n}$, where $v_1, \dots, v_n \in \mathbb{R}^2$, and $n \in \mathbb{N}$ can be different for each star.
Intuitively, it seems to me that each point $x \in G$ of the topological graph should have a neighborhood homeomorphic to a star. Is this intuition correct? How would I go about proving this?
Here's my attempt:
Let $x\in G$ be given. If $x$ lies within the interior of an edge $e$, say $x=e(t)$, then because $e$ is an embedding we may choose some open neighbourhood $U=(t-\epsilon,t+\epsilon)\subseteq[0,1]$ about $t$ such that $e(U)$ intersects $G$ in only the interior of $e$ (we could use the compactness of $G$ to do this). Now define a map $g:e(U)\rightarrow\mathbb{R}\times\{0\}$ by $e(t+\eta)=(\eta,0)$ where $\eta\in(-\epsilon,\epsilon)$. That this is a homeomorphism is (seems) clear. Then define $\vec{v}_{1}=[-\epsilon,0]$ and $\vec{v}_{2}=[\epsilon,0]$. We then have that the star $s_{\vec{v}_{1}}\cup s_{\vec{v}_{2}}$ is homeomorphic to an open neighbourhood of $x$ in $G$.
Generalizing this to vertices of $G$ is similar, only instead of two vectors you will have finitely many. Just map the various segments to some vectors of "rational slope".