In the proof that fiber bundles have the homotopy lifting property for disks (i.e. it's a Serre fibration), Hatcher writes:
I know how to use compactness to show we can subdivide $I^n\times I$ enough to make each $C\times I_j$ go to a single $U_\alpha$ by taking the standard metric in $I^n\times I$ and using the Lebesgue number $\delta$ of the cover $G^{-1}(U_\alpha)$: halve the cube as many times as needed to get each subcube with diameter smaller than $\delta$, i.e. $N$ such that $\sqrt{n+1}/2^N<\delta$.
Question: Is there another trick to do this without using a metric, just some slick point-set topology argument?
I read Is there any generalization of Lebesgue's number lemma?, but by Hatcher's writing there seems to be a simpler point-set topology argument he's referring to.
There's probably a bunch of ways to go about this and I think it's an all roads leads to Rome scenario. I can't speak for what exactly Hatcher has envisioned, but here's an argument: the $g^{-1}(U_{\alpha})$ form an open cover of $I^n\times I$, which has a finite subcover by compactness and I will assume the indices to run through a finite subset of the original indices realizing such a finite subcover from now on. Proceed by contradiction. If the assertion were not true, we could find a cube $C_k\subseteq I^{n+1}$ (for simplicities sake, I will not refer to this cube as product of a cube and an interval as is done in the text) in the $k$-th iterated subdivision along with points $x_{k,\alpha}\in C_k\setminus g^{-1}(U_{\alpha})$ for all $\alpha$ and each $k$. By compactness of $I^{n+1}$, we can find a subsequence $x_{k_n,\alpha}$ s.t. $x_{k_n,\alpha}\rightarrow x_{\alpha}$ for each $\alpha$. Now, on one hand, since the $g^{-1}(U_{\alpha})$ are open, we have $x_{\alpha}\not\in g^{-1}(U_{\alpha})$ for each $\alpha$. On the other hand, $\mathrm{diam}(C_k)\rightarrow0$, so the $x_{\alpha}$ are actually all equal (you might object I'm calling upon the metric here, but I'm only doing this insofar as I have to, since the Euclidean topology is defined in terms of a metric). Since the $g^{-1}(U_{\alpha})$ cover $I^{n+1}$, those two assertions yield a contradiction.