= Solution
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 is
$$
\boxed{f\simeq g,\qquad g(S^n)\subseteq X_n.}
$$
There 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 $0<r'<r''<1$ and a smooth cutoff $\chi$ supported in $F^{-1}(B_{r''})$ and equal to one near the compact set $F^{-1}(\overline B_{r'})$. By <smooth approximation of maps into Euclidean space>, choose a smooth $h:U\to\mathbb R^d$ with $|h-F|<\delta$, where $\delta<\min(r'/2,1-r'')$. Replace $F$ by
$$
G=F+\chi(h-F).
$$
The straight-line interpolation is a <homotopy> supported away from the boundary of the cell and stays inside that cell. Where $\chi\ne1$, we have $|F|>r'$ and hence $|G|>r'/2$. By the <Sard theorem>, choose $z\in B_{r'/4}$ outside $h(U)$. Where $\chi=1$ we have $G=h$, and elsewhere $|G|>r'/2$, 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.
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