(Reading Difficulty: 4/5)
(Prerequisites: Understand the title)

I don't like the standard proof, so here's something I thought of.

THEOREM: A set is compact iff it is closed and bounded.

Proof.

That compact sets are closed and bounded is not hard. The bigger problem is the other direction. Let us suppose is closed and bounded. Take an open cover .

Step 1: Reduction to the countable case

(Motivation: If we truly can find a finite subcover for any , then surely we should be able to do it in the case that is countable...)

Note: This entire step is skippable by citing that is a Lindelof space.

Consider the countable family of open balls . The big claim is that
covers . Proof

Now for each ball we find the open set that it is contained in. Clearly is a countable open subcover for .

Step 2: Win instantly

(Motivation: Seems kinda hard to constructively generate the finite subcover, so let's go by contradiction instead.)

We claim that admits a finite subcover for . This is equivalent to finding so large that . To prove this, suppose we could not find such . Then for all there is that is not contained in any of the open sets .

is bounded so is bounded so it has an accumulation point by Bolzano-Weierstrass (...is infinite?). (Motivator: Since each is defined to be a point that is not in a lot of the , we can guess that isn't in any . Indeed...)

CLAIM: is not contained in the open cover . In particular .

This is because if for some , then there is a small ball . But is an accumulation point of , and so in particular I can find infinitely many . But that's bogus: can contain at most the points (because by definition of the sequence , we have that ), so can only have finitely many .

CLAIM:

This is because is closed... so it contains all of its accumulation points.

These two claims are in contradiction.