A Universe of Sorts

Theorem: closed bounded subset of $\R^n$ is compact
We will prove it for $\R$ and leave the obvious generalization to the reader.
Key idea: recall that for metric spaces, compactness and sequential compactness are equivalent, so the proof must follow some ideas from Bolzano Weirstrass (sequence in closed bounded set has convergent subsequence).
Recall that that proof goes by bisection, so let's try to bisect some stuff!
Also recall why this fails in infinite dimensions: you can bisect repeatedly in "all directions" and get volume (measure) to zero, without actually controlling the cardinality. There is no theorem that says "measure 0 = single point". So, the proof must rely on finite dimension and "trapping" a point.
Take an interval, say $[0, 1]$ and take a cover $\mathcal C$ . We want to extract a finite subcover.
For now, suppose that the cover is made up only of open balls $B(x, \epsilon)$ . We can always reduce a cover to a cover of open balls --- For each point $p \in X$ which is covered by $U_p$ , take an open ball $B_p \equiv B(p, \epsilon_p) \subseteq U$ . A finite subcover of the open balls $\{ B_p \}$ tells us which $U_p$ to pick from the original cover.
Thus, we shall now assume that $C$ is only made up of epsilon balls of the form $C \equiv \{ B(p, \epsilon_p) \}$ .
If $C$ has a finite subcover, we are done.
Suppose $C$ has no finite subcover. We will show that this leads to a contradiction.
Since we have no finite subcover, it must be the case that at $I_0$ , there are an infinite number of balls $\{ B \}$ . Call this cover of infinite balls $C_0$ .
Now, let the interval $I_1$ be whichever of $[0, 1/2]$ or $[1/2, 1]$ that has infinitely many balls from $C_0$ . One of the two intervals must have infinite many balls from $C_0$ , for otherwise $C_0$ would be finite, a contradiction. Let $C_1$ be the cover of $I_1$ by taking balls from $C_0$ that lie in $I_1$ .
Repeat the above for $I_1$ . This gives us a sequence of nested intervals $\dots \subset I_2 \subset I_1 \subset I_0$ , as well as nested covers $\dots \subset C_2 \subset C_1 \subset C_0$ .
For each $i$ , pick any epsilon ball $B_i(p_i, \epsilon_i) \in C_i$ . This gives us a sequence of centers of balls $\{ p_i \}$ . These centers must have a coverging subsequence $\{ q_i \}$ (by bolzano weirstrass) which converges to a limit point $L$ .
Take the ball $B_L \equiv (L, \epsilon_L) \in C$ which covers the limit point $L$ .
Since the sequence $\{ q_i \}$ is cauchy, for $\epsilon_L$ , there must exist a natural $N$ such that for all $n \geq N$ , the points $\{ q_n : n \geq N \} \subseteq B_L$ .
Thus, we only have finitely many points, $q_{\leq n}$ to cover. Cover each of these by their own ball.
We have thus successfully found a covering for the full sequence!

§ Heine Borel