Proving a set is compact without Heine-Borel

Question

So often in real analysis and topology you're given the definition of compactness and then you're shown examples that include sets that are not closed nor bounded. So basically, these are nonexamples to the compactness definition. They always give you sets like $[0,1)$ or $(-1,1)$ and then you find open covers that have no finite subcovers, but I have very scarcely seen examples that illustrate when sets do have finite subcovers. So, I was wondering if there was a way to show that every compact set (closed and bounded here) has a finite subcover, but maybe I would be interested in seeing it with examples on the real line.

So let's take $X=[a,b] \in \mathbb{R}$. First, we know $\overline X = X$ and we know that $X$ is bounded. How can I create an arbitrary open cover from any interval on the real line and show that it has a finite subcover?

I mean, since $X$ is closed and bounded, we already "know" that $X$ is compact, but suppose we didn't have the Heine-Borel Theorem to help us. How could we show that $[a,b]$ is compact without using Heine-Borel?

Answer 1

$[a,b]$ is compact because it has the order topology and every subset of $[a,b]$ has a supremum.

The fastest proof is by Alexander's subbase theorem, using the subbase $$\mathcal{S} = \left\{[a,x): x \in [a,b]\right\} \cup \left\{(x,b]: x \in [a,b]\right\}$$

Let $\mathcal{U} \subseteq \mathcal{S}$ be an open cover of $[a,b]$.

To cover $a$, we need at least one set of the form $[a,x)$, $x \in (a,b]$ to be in $\mathcal{U}$. So $A = \{x\mid [a,x) \in \mathcal{U}\}$ is non-empty and has a supremum $c \in [a,b]$. This $c$ is not covered by any set of the form $[a,x) \in \mathcal{U}$ (because it's larger than all those $x$ by definition) so it must be covered by some $c \in (c',b]$ with $(c',b] \in \mathcal{U}$. But then $c' < c$ so there must be some $d'$ such that $[a,d') \in \mathcal{U}$ and $d' > c'$ as $c=\sup A$. But then $[a,d') \cup (c',b] = [a,b]$ form a finite subcover of $\mathcal{U}$ and we're done.

No Heine-Borel needed at all. Just the completeness of $\Bbb R$ which does it.