Metacompact Countably Compact Space is Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a countably compact space which is also metacompact.

Then $T$ is compact.

Proof
From the definition of countably compact space, every countable open cover of $S$ has a finite subcover.

Let $T = \left({S, \tau}\right)$ be a countably compact space which is also metacompact.

Let $\mathcal U_\alpha$ be an open cover of $S$.

Then let $\mathcal V_\beta$ be the open refinement which is point finite, guaranteed by its metacompactness.

Let $x \in S$.

We have that $x$ is an element of only a finite number of the elements of $\mathcal V_\beta$, as $\mathcal V_\beta$ is point finite.

Consider all the subcovers of $\mathcal V_\beta$, and order them by subset.

Consider any chain $\mathcal C$ of such subcovers.

Suppose $x$ is not in the intersection of $\mathcal C$.

Then it would fail to be covered by one of the elements of $\mathcal C$.

This would be a contradiction of the fact that every element of $\mathcal C$ is a cover of $\mathcal V_\beta$.

So the intersection of a chain of such subcovers is itself a subcover.

Hence it must be that $\mathcal V$ has a minimal subcover, $\mathcal V_\gamma$, say.

Now each element $V_\gamma$ of $\mathcal V_\gamma$ must itself contain a unique element $x_\gamma \in V_\gamma$ which belongs to no element of $\mathcal V_\gamma$.

This is because $\mathcal V_\gamma$ is minimal.

If $\mathcal V_\gamma$ were infinite, the set of all $\left\{{x_\gamma}\right\}$ would be an infinite set without an $\omega$-accumulation point.

This can not be the case.

Thus $\mathcal V_\gamma$ is a finite cover.

So we have constructed a finite subcover for $\mathcal U_\alpha$, demonstrating that $T$ is compact.