Showing homotopy equivalence of identity map from graph topology to metric topology on the same set To elaborate on the question:
Consider set X equipped with 1-dimensional CW complex topology ($\mathcal{T_{CW}}$) which we can think essentially as graph. Edges are thought of as intervals $[0,1]$.
Exclude possibilities of:

*

*vertices connected by more than 1 edge

*loops (vertices connected to themselves by an edge)

$V = X^{(0)}$ is a set of vertices.
$M = $ the set of all $f: V \to \mathbb{R}$ for which there exists a finite subset $\{v_1, ..., v_n\} \subset V$ such that for all $v \in V - \{v_1, ..., v_n\}$ we have $f(v)=0$
Consider the injection $g: X \to M$ where $g_{|V}$ is defined as $g(v): V \to M$ by
$$g(v)(u) = \begin{cases} 0 & v \not= u \\ 1 & v=u \end{cases} 
$$
For an edge $E$ thought of as $[0,1]$, $g_{|E}$ is defined using the convex map $g(t) = (1-t)g(v) + tg(u)$.
Then $d(f, h) = \sum_{v \in V} |f(v) - h(v)| \ \forall f,h \in R$ is a metric on $g(X)$ denoted $\mathcal{T_{d}}$.
The question is how do you prove that $Id: (X, \mathcal{T_{CW}}) \to (X, \mathcal{T_{d}})$ is indeed a homotopy equivalence?
I mean it's easy to see that open sets in $X$ are open in metric topology on the same set. Then you can show continuity of identity map from graph to metric topology on the set $X$, but not the other way. For instance, if you take infinite number of vertices each of length $\frac{1}{n}$. But to show homotopy equivalence I struggle to come up with formal proof as the identity map is not homeomorphism so you need a homotopy inverse?
 A: You can define a homotopy inverse $J : (X, \mathcal{T_{d}}) \to (X, \mathcal{T_{CW}})$ to the map $Id: (X, \mathcal{T_{CW}}) \to (X, \mathcal{T_{d}})$ as follows.
First I'll note that in the metric you have defined, each edge $E$ is isometric to $[0,2]$. For each vertex $v$ let $N(v)$ be the radius $1/2$ neighborhood of $v$ in the topology $\mathcal{T_{d}}$; it follows that for each edge $E \approx [0,1]$ incident to $v$, if $0$ corresponds to $v$ then the intersection of $N(v)$ with $E$ is $[0,1/2)$, whereas if $1$ corresponds to $v$ then the intersection of $N(v)$ with $E$ is $(3/2,2]$.
Now define $J$ by the following formula. For, for each vertex $v$ set $J \mid N(v)$ to be the constant map with value $v$. Consider next each edge $E \approx [0,1]$, so $0,1$ correspond to two vertices, and $J$ is already defined on $[0,1/2] \cup [3/2,1]$. On the remaining portion $[1/2,3/2]$ define $J(t) = 2t-1$, $1/2 \le t \le 3/2$.
Continuity of $J$ is pretty easy to check.
Set theoretically the two maps $Id \circ J$ and $J \circ Id$ are both equal to $J$, and there is an obvious homotopy $H : X \times [0,1] \to X$ between $Id$ and $J$ which is continuous in both of the product topologies $(X, \mathcal{T_{d}}) \times [0,1]$ and $(X, \mathcal{T_{CW}}) \times [0,1]$.
