$\newcommand\met{\mathrm{met}}$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.



Making CW-complexes metrizable
The required metric topology on $X$ does exist. This is a consequence of Theorem 2.1 in the paper



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> Robert Cauty, *Rétractions dans les espaces stratifiables*, Bulletin de la Société Mathématique de France, **102**, (1974), 129-149.
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Actually, Cauty's statement is far more general than what is being required. I'll record it in the present context and postpone discussion for afterwards.



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> **Theorem (Cauty)**: Let $X$ be a CW complex. Then there exists a continuous metric $d$ on $X$ such that;
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> * $X\_{Met}=(X,d)$ is an ANR.
> * The identity $i:X\rightarrow X\_{Met}$ is a homotopy equivalence.
> * For any given compact $K\subseteq X$, a homotopy inverse $j:X\_{Met}\rightarrow X$ to $i$ can be found so that $i\circ j\simeq id\_{X\_{met}}$ and $j\circ i\simeq id\_X$ by homotopies fixing $K$ pointwise at all times. $\quad\blacksquare$
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There are no restrictions on the dimension of $X$ or its number of cells. A metric is continuous if and only if it comes from a weaker metric topology.


Now, as mentioned above, Cauty's actual statement is far more general.



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> *The theorem holds verbatim when $X$ is any ANR for stratifiable spaces*.
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The term *stratifiable* is meant in the sense of [Borges](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-17/issue-1/On-stratifiable-spaces/pjm/1102994724.full), who established already that a stratifiable space belongs to ANR(stratifiable) if and only if it belongs to ANE(stratifiable). Ceder had previously established the result that all CW complexes are stratifiable, and Cauty subsequently showed that all CW complexes belong to ANR(stratifiable).


Ignoring all references to stratifiability, we can at the very least use the extra generality to extend the theorem stated above to include the cases $(i)$ $X$ is the product of finitely many CW complexes, $(ii)$ $X$ is an open subset of a CW complex, $(iii)$ $X$ is a closed subset of a CW complex $Y$ whose inclusion $X\subseteq Y$ is a cofibration, $(iv)$ $X$ is the loop space of a CW complex.


Finally, let us comment on the last listed property regarding the compact $K$. Any compact subset $K\subseteq X$ has the same topology as its image in $X\_{Met}$. The statement says that $i:X\rightarrow X\_{Met}$ is a homotopy equivalence under $K$. This is clearly true when $K$ is a finite subcomplex of $X$, since then its inclusions $K\subseteq X$ and $K\subseteq X\_{Met}$ are cofibrations (the latter following since $K$ is an ANR for metric spaces). That it should remain true for any $K$ (say a Cantor set or compact fractal) is quite surprising!


