Compact implies countably compact. But I'm having difficulty with proving the other direciton. My work:
Suppose $(X, \mathcal{O})$ is second countable and countably compact. By definition, there is a collection of open sets $\{B_\alpha\}_{a\in I}$, with $I$ countable, such that any open set in $X$ can be written as a union of $B_\alpha$'s. Suppose $\{E_\alpha\}_{\alpha\in \Omega}$ is an open cover of $X.$ Then, for each $\alpha\in \Omega$,
$$E_\alpha= \bigcup_{i\in I_\alpha} B_i$$
for some $I_\alpha \subset I.$ Thus,
$$X = \bigcup_{\alpha \in \Omega} \bigcup_{i \in I_\alpha} B_i.$$
This is a countable open cover of $X$, so there must exist finitely many $B_i$'s which cover $X,$ but I'm struggling to use this to say anything about the arbitrary open cover.
Best Answer
Let $\mathcal{B}_n, n \in \Bbb N$ be any countable base for $X$, a countably compact space.
If $\mathcal{U}$ is then any open cover for $X$, define $N = \{n \in \Bbb N: \exists U(n) \in \mathcal{U}: B_n \subseteq U(n)\}$, which is a countable set (subset of $\Bbb N$) and $\mathcal{U}':=\{U(n) \mid n \in N\}$ is a countable subcover of $\mathcal{U}$ (if $x \in X$, there is some $U \in \mathcal{U}$ so that $x \in U$ so there is some $m\in \Bbb N$ so that $x \in B_m \subseteq U$ and then it's clear that $x \in U(m)$ and $m \in N$ by definition, so $x$ is covered by $\mathcal{U}'$) and so has a finite subcover by countable compactness. So $X$ is compact.