TITLE: Making CW-complexes metrizable
QUESTION [11 upvotes]: $\newcommand\met{\mathrm{met}}$It is a basic topological fact that CW-complexes aren't typically metrizable (they must satisfy a certain local finiteness condition) and the quotient topology is to blame.
Question: Suppose $X$ is a CW-complex (possibly with countably many cells and maybe even of finite dimension). Is it possible weaken the topology of $X$ to construct another space $X_{\met}$ (with the same underlying set), so that the continuous identity function $X\to X_{\met}$ is a homotopy equivalence?
Update: I will clarify (now much later) that this question has an affirmative answer for simplicial complexes. Given an arbitrary simplicial complex $K$, we have $|K|$, which has the weak topology and is not always metrizable. However, you can give the underlying set of $|K|$ a metrizable topology to form the "metric simplicial complex" $|K|_m$. The identity $|K|\to |K|_m$ is continuous and is a homotopy equivalence. A nice proof can be found in Segal and Mardesic's book Shape Theory in Appendix, $\S 1.3$, Theorem 10. As Sergey Melikhov nicely points out in his answer, the same is true for regular CW-complexes, which include simplicial complexes. Using this result, it follows that every CW-complex is homotopy equivalent to some metric space. However, my question is a bit more specific.

REPLY [3 votes]: I had trouble finding a reference online for a non-metrizable CW complex, so I figured it might be nice to record that here. Let $X = \bigvee_{n=1}^\infty S^1$ be the infinite bouquet of circles, considered as a CW complex. I claim that this space is not metrizable. An important fact about CW topology that we can exploit is that any open neighborhood of the basepoint $x_0$ intersects the interior of every 1-cell of this cell complex.
Now suppose $X$ is metrizable, i.e. has a metric $d$ which induces the same topology as that of the cell structure. Any open ball centered at $x_0$ intersects the interior of every 1-cell. So we can pick a sequence $(x_n)_{n=1}^\infty$ such that $x_n$ lies in the interior of the $n$-th 1-cell and $d(x_0,x_n)<1/n$. Then $\lim x_n = x_0$ by construction. On the other hand, the set $X-\{x_1,x_2,\dotsc\}$ is an open neighborhood of $x_0$ since it is open in each cell of the complex. Thus $x_0$ is separated from the sequence $(x_n)_{n=1}^\infty$, a contradiction. $\quad \square$