In Introduction to Topological Manifolds by J. M. Lee this result proved with a use of an axiom of choice. In my words:
If $K$ is a compact subset of CW-complex $(X,\mathcal{E},\varphi)$ in question, let $\mathcal{K} = \{ e \in \mathcal{E} : e \cap K \neq \emptyset \}$. Assume contrapositive, which means that $|\mathcal{K}| > \infty$. Then, choose a collection of points $D =\{d_e \in e, e \in \mathcal{K}\}$. Then by the property of cellular structure, for each cell $e \in \mathcal{E}, |e \cap D|$ is either $0$ or $1$, so by preceding lemma the set $D$ is closed and discrete. Compact sets cant' contain infinite closed discrete subsets, a contradiction.
My only complain about this proof is that 'choose' is a direct use of the axiom of choice. But this seems to be a very basic property of CW-complex, hence the question: do we actually need axiom of choice here?
Personally I see here two possibilities:
First one is less likely, that this statement actually requires AC. In this case, in $\neg\mathrm{AC}$-world it would be possible to construct a CW-complex with a compact subset that can't be contained in any finite subcomplex. And I find this possibility as pretty exciting!
The more likely possibility is that there is that there is another proof, which uses a weaker form of AC $\mathbf{A}$, say ultrafilter lemma. I'm thinking ultrafilter lemma, because it's natural to use ultrafilters and compact sets together. I haven't figured how to use it here yet. But even in this case it might be possible to consider $\neg \mathbf{A}$-universe and construct a CW-complex with such a weird property.
p.s.
I don't think that many people thought about this question, because (introductory) algebraic geometry appeals towards visual thinking, while discussions about axioms appeals towards something inverse.
Best Answer
I do not have access to Lee's book, but the notation $(X,\mathcal{E},\varphi)$ suggests that the characteristic maps of the cells are part of the structure.
In that case we do not need AC. In fact, for each cell $e = e^n_\alpha$ we have an associated characteristic map $\varphi_e : D^n \to X$. Then take $d_e = \varphi_e(0)$.
If the characteristic maps do not belong to the structure and it is only required that for each cell $e$ there exists a characteristic map, then we are in the situation of your question and in my opinion AC must be used. However, the axiom of countable choice is sufficient. To see this, note that if $\mathcal K$ is infinite, we can get a countably infinite set $D$ of points which is closed and discrete.
I recommend to have a look into
On p. 31 the authors discuss the use of AC in the context of proving that a CW-complex is paracompact.