Second-Countable Space is Compact iff Countably Compact

Theorem
Let $T = \struct {S, \tau}$ be a second-countable space.

Then $T$ is compact $T$ 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 $\BB$ for $\tau$.

Let $\CC$ be an open cover for $S$.

Define:
 * $\CC' = \set {B \in \BB: \exists C \in \CC: B \subseteq C}$

Then $\CC' \subseteq \BB$, and so $\CC'$ is countable.

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

From Equivalence of Definitions of Analytic Basis:
 * $\forall C \in \CC: \forall x \in C: \exists B \in \CC': x \in B$

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

Hence, by Union is Smallest Superset:
 * $\ds S \subseteq \bigcup \CC \subseteq \bigcup \CC'$

Since $\subseteq$ is a transitive relation, it follows that $\CC'$ is an open cover for $S$.

Since $T$ is countably compact, there exists a finite subcover $\FF$ of $\CC'$ for $S$.

By the Principle of Finite Choice, there exists an indexed family $\family {C_B}_{B \mathop \in \FF}$ such that:
 * $\forall B \in \FF: B \subseteq C_B \in \CC$

Note that $\set {C_B: B \in \FF}$ is a finite subcover of $\CC$ for $X$.

Hence, $T$ is compact.