I'm confused by the proof of Heine-Borel. The proof I'm referring to is the following:
Assume that there is a closed and bounded set $X$ of size $s$ which isn't compact. Then we can split $X$ into two sets and at least one of the sets is not covered by a finite subcover. WLOG we can say one of the sets (call it $X_1$) is not covered by a finite subcover. We can continue the subdivision of $X_1$ and so on, getting a sequence of sets $\{X, X_1, X_2, X_3, ...\}$, and we know by the nested interval theorem that this sequence contains a point, say $x_0$. Then we can take the open set containing $x_0$, call it $C$ and since the sequence of sets leads to arbitrarily small sets, we eventually end up with some set $X_n$ of arbitrarily small size which is covered by $C$. Contradiction, therefore $X$ is compact.
To me, this seems to be saying something about the sum of open sets - I'm envisioning this method of subdivision saying that $\sum_n{(b-a)/2^n} = b-a$, and you don't need $n$ to go to infinity because there's an open set around $b$ with a fixed radius. However, in my mind that only holds true if you assume only one of the sets in the subdivision is not covered by a finite subcover - what if both of the sets in the subdivision is assumed to not be covered by a finite subcover at every single point? Then wouldn't there be a possibly infinite amount of points $x_0, x_1...$ (and therefore an infinite subcover). This could hold true if the open sets around the various points $x_0, x_1...$ could have increasingly smaller and smaller radii? If the radii decreased faster than exponentially then why wouldn't the number of subdivisions (and thus, covers) go to infinity as well?
Essentially I can see how you can take care of one sequence with this proof method, but I don't see how you can take care of an exponentially growing amount of sequences.