Second-Countable Space is Compact iff Countably Compact

Theorem
Let $T = \left({X, \vartheta}\right)$ be a second-countable space.

Then $T$ is compact iff $T$ is countably compact.

Proof
Let $T = \left({X, \vartheta}\right)$ be a second-countable space.

We have that a Compact Space is Countably Compact whether the space is second-countable or not.

So what we need to show is that if $T = \left({X, \vartheta}\right)$ is countably compact it follows that it is also compact.

So, let $T = \left({X, \vartheta}\right)$ be a countably compact.

As $T$ is second-countable, its topology has a countable basis.

Let this basis be $\mathcal B$.

Let $\mathcal C$ be an open cover of $X$.

As $T$ is second-countable, $\mathcal C$ can be expressed as the union $\mathbb S$ of sets of $\mathcal B$.

As $\mathcal B$ is countable, so is any such subset, and so $\mathbb S$ is a countable open cover of $X$.

As $T$ is countably compact, $\mathbb S$ has a finite subcover.

Thus we have shown that $\mathcal C$ has a finite subcover.

Hence $T$ is compact.