Second-Countable Space is Compact iff Countably Compact

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

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

Necessary Condition
We have that a compact space is countably compact (whether the space is second-countable or not).

Sufficient Condition
Let $T$ be countably compact.

Since $T$ is second-countable, we can choose a countable (analytic) basis $\mathcal B$ for $\tau$.

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

Define:
 * $\mathcal C' = \left\{{B \in \mathcal B: \exists C \in \mathcal C: B \subseteq C}\right\}$

Then $\mathcal C' \subseteq \mathcal B$, and so $\mathcal C'$ is countable.

By the definition of an analytic basis, we have $\mathcal B \subseteq \tau$; since $\subseteq$ is a transitive relation, it therefore follows that $\mathcal C' \subseteq \tau$.

From Equivalent Definitions of Analytic Basis, we can conclude that:
 * $\forall C \in \mathcal C: \forall x \in C: \exists B \in \mathcal C': x \in B$

That is, by the definition of set union:
 * $\displaystyle \forall C \in \mathcal C: C \subseteq \bigcup \mathcal C'$

Hence, by Union Smallest: General Result:
 * $\displaystyle X \subseteq \bigcup \mathcal C \subseteq \bigcup \mathcal C'$

Since $\subseteq$ is a transitive relation, it follows that $\mathcal C'$ is an open cover for $X$.

Since $T$ is countably compact, there exists a finite subcover $\mathcal F$ of $\mathcal C'$ for $X$.

Since the finite cartesian product of non-empty sets is non-empty, there exists a family $\left\langle{C_B}\right\rangle_{B \in \mathcal F}$ such that:
 * $\forall B \in \mathcal F: B \subseteq C_B \in \mathcal C$

Note that $\left\{{C_B: B \in \mathcal F}\right\}$ is a finite subcover of $\mathcal C$ for $X$.

Hence, $T$ is compact.