1. The problem statement, all variables and given/known data
The lemma to prove is that "If [-R,R]^{n-1} is compact, then [-R,R]^n is too.

To help us, we have two other lemmas already proven:

L1: "[a,b] is compact."

L2: "If A is R^n is compact and x_0 is in R^m, then A x {x_0} is compact."

3. The attempt at a solution I found a proof that is substantially simpler than the one provided by the book. Usually, this is not a good sign!

I say, let [tex]\{U_i\}_{i\in I}[/tex] be an open cover for [-R,R]^n. For all x in [-R,R], [-R,R]^{n-1} x {x} is compact (by L2), so we can find a finite subcover, say [tex]\{U_{i^{(x)}_k}\}_{1\leq k\leq N_x}[/tex]. Let [tex]O_x=\cup_{k=1}^{N_x}U_{i^{(x)}_k}[/tex]. We have that O_x is open and [-R,R]^{n-1} x {x} [tex]\subset[/tex] O_x. Next, observe that the familly [tex]\{O_x\}_{x\in [-R,R]}[/tex] covers [-R,R]^n. As in the proof of L2,

By construction, the resulting sub-family, say [tex]\{O_{x_j}\}_{1\leq j\leq M} [/tex] is a finite cover of [-R,R]^n. Therefor, [tex]\{U_{i^{(x_j)}_1},...,U_{i^{(x_j)}_{N_{x_j}}}\}_{1\leq j \leq M}[/tex] is a finite subcover of [-R,R]^n that has been extracted from our original cover [tex]\{U_i\}_{i\in I}[/tex].

Is it not so that since {O_x} covers [-R,R] x [-R,R], the projection of the last coordinates of the O_x on the real line covers [-R,R] with open sets? we can then extract a finite subcover of these projections, and after we pullback again, the whole O_x's corresponding to the finite subcover actually covers [-R,R]^n because by construction, each O_x covers [-R,R]^{n-1}.

How is the situation different here than in my other post?

I think what you're trying to do, starting at the quote, is something like this:

To every x, define A_x to be the open subset of [-R,R] obtained by projecting the sets in O_x onto the last coordinate. Then finitely many of the A_x cover [-R,R], and taking the union of the corresponding O_x gives a finite subcover of U_i.

If this is what you meant, it's not right. It seems like you're assuming that if the projection of the O_x is A_x (an open subset of [-R,R] containing x), then O_x covers not only [-R,R]^{n-1}x{x}, but in fact all of [-R,R]^{n-1} x A_x.

But A_x gives, in a sense, the "tallest" "cross-section" of the O_x. But O_x will only cover [-R,R]^{n-1} x B_x where B_x is the shortest cross-section of the O_x. For a general set, there might not be such a B_x, since you may have a set in the cover like {(x,y)| |y|<1/|x|}. But here, using the compactness of [-R,R]^{n-1}, you should be able to find such a set.

I think I misread part of your post. I still can't quite decide if you need to add anything to your reasoning - I will ponder some more. But I'm beginning to think it is correct. Indeed, I'd now go so far as to say it is obviously correct and I was being an idiot.

Well, in any case, StatusX's post showed me why what I thought I was doing was not correct. I was having difficulty with this proof and Lemma 2's proof because I obstinately tried to visualize the thing in n dimensions because I am always afraid that if I work with R² or R³, I might do something that is not legal in an arbitrary n.

In any case, it was easy to see in R² what bit of thinking I was doing wrong, but the fact that [-R,R]^{n-1} is compact assures us, (by defining the right function and taking it's inf) that we can find an epsilon_y>0 such that O_y covers [-R,R]^{n-1} x (y-epsilon_y, y+epsilon_y). And the fact that [-R,R] is compact (see L1) assures usagain that inf{epsilon_y: y in [-R,R]} is stricly greater than zero. From there, it's just a technicality to construct a subcover.

Overall, my proof is only a semi-success, because it leans on the fact that a function defined on a compact takes its infimum, a result not yet seen. And if one formalizes, the idea hinted at by StatusX and that I described just above, the result is a shower of symbols detailing technical details and in the end, the proof is not any easier to read than the one provided by the book. Though it is more intuitive!