$\newcommand\met{\mathrm{met}}$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.
Best Answer
This addresses the modified question in Jeremy's comments, on keeping the preferred CW-structure.
If the CW complex happens to be regular and PL (i.e. the attaching maps are injective and piecewise-linear), its barycentric subdivision is a simplicial complex (namely, the order complex of the poset of nonempty faces of the CW complex), which can be endowed with the usual barycentric metric. The identity map will then be a homotopy equivalence (proofs can be found in some old textbooks, including the Appendix of Dold's Algebraic topology, or "Theory of retracts" by Hu Sze-Tsen).
For a general (countable) CW complex, one can inductively homotop the attaching maps of $(n+1)$-cells by a homotopy with values in the $n$-skeleton so that the modified CW complex $K$ admits a barycentric subdivision $K'$ that is a regular simplicial set, in the sense that cells of $K$ are identified with the unions of simplices of $K'$ whose first vertex is a fixed $0$-simplex of $K'$. (Regular means that the representing map of every non-degenerate simplex only makes identifications along the last facet of the simplex.) The geometric realization of a regular simplicial set is a regular CW-complex, so the previous construction applies. (In more detail, the order complex $K''$ of the poset of nonempty nondegenerage simplices of $K'$, ordered by inclusion, is a simplicial complex.) An enlightening overview of subdivisions of simplicial sets can be found here.
The homotopies of attaching maps can be constructed using Brouwer's simplicial approximation theorem, which implies that any continuous map $|L|\to |X|$ between geometric realizations of finite simplicial sets is homotopic, upon precomposing with the geometric realization of an iterate $L^{(n)}\to L$ of the last vertex map $L'\to L$, to the geometric realization of a morphism $f:L^{(n)}\to X$ of simplicial sets (see Corollary 3.2 here). Here $L$ is any triangulation $S^n$ by a non-singular simplicial set (i.e. a subcomplex of the order complex of a poset), and $X$ is the $n$-skeleton of $K'$, which is a regular simplicial set. Then the mapping cone of $f$ is again a regular simplical set.
In light of the combinatorial view of regular PL CW complexes, one could try to homotop the attaching maps of a general CW complex so as to achieve a more rigid combinatorial structure of the simplicial set $K'$. However, a homotopy class is not generally representable by a non-degenerate map (in the sense of collapsing no simplices). Because of this, $K'$ cannot be generally chosen to be the nerve of a category, nor even a quasi-category.