Exercise 2.26

We know that $X$ is separable by ex. 24, and that it has a countable base by ex. 23. Since any open set is a union of base sets, we can reduce any open cover to a countable subcover. Let $G=\{G_n\}$ be such a subcover. Assume $G$ has no finite subcover, so that $F_n={(G_1\cup \cdots \cup G_n)}^c$ is non-empty for all $n$. Pick $x_n\in F_n$. They constitute an infinite subset, and therefore have a limit point $x$. This point is in $G_k$ for some $k$. But then $F_k^c$ is an open set around $x$ that contains only a finite number of the $x_n$, contradicting $x$ being a limit point. Any open cover of $X$ must therefore have a finite subcover, so $X$ is compact.