The Compactness Theorem is equivalent to the compactness of the Stone space of the Lindenbaum–Tarski algebra of the first-order language $L$. (This is also the space of $0$-types over the empty theory.)
A point in the Stone space $S_L$ is a complete theory $T$ in the language $L$. That is, $T$ is a set of sentences of $L$ which is closed under logical deduction and contains exactly one of $\sigma$ or $\lnot\sigma$ for every sentence $\sigma$ of the language. The topology on the set of types has for basis the open sets $U(\sigma) = \{T:\sigma\in T\}$ for every sentence $\sigma$ of $L$. Note that these are all clopen sets since $U(\lnot\sigma)$ is complementary to $U(\sigma)$.
To see how the Compactness Theorem implies the compactness of $S_L$, suppose the basic open sets $U(\sigma_i)$, $i\in I$, form a cover of $S_L$. This means that every complete theory $T$ contains at least one of the sentences $\sigma_i$. I claim that this cover has a finite subcover. If not, then the set $\{\lnot\sigma_i:i\in I\}$ is finitely consistent. By the Compactness Theorem, the set consistent and hence (by Zorn's Lemma) is contained in a maximally consistent set $T$. This theory $T$ is a point of the Stone space which is not contained in any $U(\sigma_i)$, which contradicts our hypothesis that the $U(\sigma_i)$, $i\in I$, form a cover of the space.
To see how the compactness of $S_L$ implies the Compactness Theorem, suppose that $\{\sigma_i:i\in I\}$ is an inconsistent set of sentences in $L$. Then $U(\lnot\sigma_i),i\in I$ forms a cover of $S_L$. This cover has a finite subcover, which corresponds to a finite inconsistent subset of $\{\sigma_i:i\in I\}$. Therefore, every inconsistent set has a finite inconsistent subset, which is the contrapositive of the Compactness Theorem.
I think that what you’re missing is that an open cover of a compact set can cover more than just that set. Let $X$ be a topological space, and let $K$ be a compact subset of $X$. A family $\mathscr{U}$ of open subsets of $X$ is an open cover of $K$ if $K\subseteq\bigcup\mathscr{U}$; it’s not required that $K=\bigcup\mathscr{U}$. You’re right that $\bigcup\mathscr{U}$, being a union of open sets, must be open in $X$, but it needn’t be equal to $K$.
For example, suppose that $X=\Bbb R$ and $K=[0,3]$; the family $\{(-1,2),(1,4)\}$ is an open cover of $[0,3]$: it’s a family of open sets, and $[0,3]\subseteq(-1,2)\cup(1,4)=(-1,4)$. And yes, $(-1,4)$ is certainly open in $\Bbb R$, but $[0,3]$ is not.
Note, by the way, that it’s not actually true that a compact subset of an arbitrary topological space is closed. For example, let $\tau$ be the cofinite topology on $\Bbb Z$: the open sets are $\varnothing$ and the sets whose complements in $\Bbb Z$ are finite. It’s a straightforward exercise to show that every subset of $\Bbb Z$ is compact in this topology, but the only closed sets are the finite ones and $\Bbb Z$ itself. Thus, for example, $\Bbb Z^+$ is a compact subset that isn’t closed.
It is true, however, that compact sets in Hausdorff spaces are closed, though a bit of work is required to establish the result.
Best Answer
Frechet oiginally coined the term in 1904 to refer to a space where every sequence had a limit point. Thus, the space was 'compact' because there was no room for the sequence to escape (my interpretation).