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A map from the $n$-sphere to a CW complex can be homotoped into its $n$-skeleton. By compact subsets of CW complexes lie in finite subcomplexes, its image lies in a finite subcomplex. For a top-dimensional cell of dimension $d>n$, use smooth approximation of maps into Euclidean space in its interior, with a cutoff supported away from the cell boundary. The Sard theorem supplies a point in a smaller interior ball omitted by the smoothed map; the cutoff is chosen so the transition region also avoids that point. Radial retraction from the omitted point pushes the image into the cell boundary and fixes all other cells. Repeating over the finitely many higher-dimensional cells proves the assertion, without assuming the initial continuous image has measure zero.

The cellular approximation theorem says that a continuous map between CW complexes is homotopic to a cellular map, meaning that it sends each $q$-skeleton into the target's $q$-skeleton. Give $S^n$, for $n>0$, its CW complex structure with one zero-cell and one $n$-cell. Its $n$-skeleton is the whole sphere, so the theorem applies and gives the requested result. For $n=0$ use the two zero-cells of $S^0$. In particular the conclusion isThere is no local-finiteness assumption on $X$.

For completeness, here is a direct proof of the needed case of cellular approximation, including the point-avoidance step. We use three standard facts, stated explicitly. First, a compact subset of a CW complex lies in a finite subcomplex. Second, a continuous map from a smooth manifold into Euclidean space has arbitrarily close smooth approximations, and smooth cutoffs exist around compact subsets of open sets. Third, the Sard theorem says the critical values of a smooth map have measure zero; when the domain dimension is smaller than the target dimension, its entire image has measure zero.

The first fact has a useful short proof. If a compact set met infinitely many open cells, choose one point in each of countably many distinct cells. Closure-finiteness makes every closed cell meet the chosen set, and every subset of it, in finitely many points. The weak topology of a CW complex then makes all those subsets closed. The chosen set is an infinite closed discrete subset of a compact space, a contradiction. Taking the closures of the finitely many cells met proves the claim.

Now $f(S^n)$ lies in a finite subcomplex $Y$. Take a top-dimensional open cell $e^d$ of $Y$ with $d>n$, and identify its interior with the open unit ball in ${\mathbb{R}}^d$ using its characteristic map. On the open set $U=f^{-1}(e^d)$, write $F:U\to B^d$ for the coordinate expression. Choose $0<r^{\prime }<r^{\prime \prime }<1$ and a smooth cutoff $\chi$ supported in $F^{-1}(B_{r^{\prime \prime }})$ and equal to one near the compact set $F^{-1}({\overline{B}}_{r^{\prime }})$. By smooth approximation of maps into Euclidean space, choose a smooth $h:U\to {\mathbb{R}}^d$ with $|h-F|<\delta$, where $\delta <\min (r^{\prime }/2,1-r^{\prime \prime })$. Replace $F$ byThe straight-line interpolation is a homotopy supported away from the boundary of the cell and stays inside that cell. Where $\chi \ne 1$, we have $|F|>r^{\prime }$ and hence $|G|>r^{\prime }/2$. By the Sard theorem, choose $z\in B_{r^{\prime }/4}$ outside $h(U)$. Where $\chi =1$ we have $G=h$, and elsewhere $|G|>r^{\prime }/2$, so the modified map avoids $z$ everywhere.

Radial retraction of $D^d\setminus \{z\}$ onto its boundary fixes that boundary. It therefore descends through the characteristic map, including its possibly noninjective boundary identifications, and fixes all other cells of $Y$. Composing gives a homotopy whose endpoint misses all of $e^d$. Remove every cell of dimension greater than $n$ in descending order. There are finitely many such cells, so their homotopies concatenate to yield $g:S^n\to Y^n\subseteq X_n$. This proves dimension reduction of sphere maps into CW skeleta. The approximation step is essential: an arbitrary continuous image of $S^n$ can fill a higher-dimensional region, so dimension alone would not supply an omitted point.