I was working on trying to give a detailed, piece by piece breakdown of the proof that the closed unit interval $[0,1]$ is compact, particularly the proof which uses the sequence of non-finitely coverable subsets to arrive at a contradiction, and have managed to do so for the most part.
Eventually, however, I realized that I had overlooked to prove one of the most important lemmas needed in the proof, that being that if you have a set $\mathbb S$ with some cover $\mathscr C$ such that $\mathbb S$ is not finitely coverable by some subcollection of $\mathscr C$, then there is some subset of $\mathbb S$ which is also not finitely coverable by any subcollection of $\mathscr C$.
My attempt at a proof is given below:
Show that if $\mathbb S$ is some set and $\mathscr C$ is some cover of $\mathbb S$ such that there is no finite subcover of $\mathscr C$ which covers $\mathbb S$, then there is some subset of $\mathbb S$ which is also not finitely coverable by a subcollection of $\mathscr C$.
Assume the opposite, that in fact there is no such subset for which a finite subcover doesn't exist, and that $\mathbb S$ is still not finitely coverable. This is the same as saying that every subset of $\mathbb S$ is finitely coverable by some finite subcollection of $\mathscr C$.
Partition $\mathbb S$ into two pieces, $P_1$ and $P_2$. Since these two subsets partition $\mathbb S$, $$P_1 \cup P_2 = \mathbb S$$ Now, since there is no subset of $\mathbb S$ which isn't finitely coverable, every subset is finitely coverable, and thus $P_1$ and $P_2$ are finitely coverable. Call the subcovering of $P_1$, $\Delta$ and the subcovering of $P_2$, $\sigma$. Taking the union of $\sigma$ and $\Delta$, we have that $\sigma \cup \Delta$ is a subcovering of $P_1 \cup P_2$. However, since $\sigma$ is finite and $\Delta$ is finite, their union is also finite, thus $\sigma \cup \Delta$ is a finite subcover of $P_1 \cup P_2$, but remember that $P_1 \cup P_2=\mathbb S$, meaning $\sigma \cup \Delta$ is a finite subcovering of $\mathbb S$, contradicting our initial assumption that $\mathbb S$ is not finitely coverable.
$Q.E.D$
I'm not particularly good at proof writing, so if someone could look this over and verify that it's correct (or incorrect), that would be much appreciated.