Countably compact if and only if countable closed sets are compact

Question

Prove that $X$ is countably compact if and only if every countable closed subset of $X$ is compact

Answer 1

(I'll assume $X$ is $T_1$; as this is an exercise in Engelking, who assumes Hausdorffness for (countably) compact spaces, this is probably not a big limitation).

Suppose $X$ is countably compact. Let $D$ be a countable closed subset of $X$. Then $D$ is also countably compact (being a closed subset of $X$) and this $D$ is then clearly compact: every open cover of $D$ has a countable open subcover (pick one set for each $d \in D$) and this subcover has a finite subcover by countable compactness.

Suppose $X$ is not countably compact. Then it has a countable cover $\{U_n: n \in \omega\}$ without a finite subcover. Then we can assume WLOG that the $U_n$ are strictly increasing (taking finite unions of initial segments and removing duplicates), so that we have $U_0 \subsetneq U_1 \subsetneq U_2 \ldots$, where all $U_i$ are open and $\neq X$. Then pick $x_n \in X\setminus U_n$, and note that $D = \{x_n: n \in \omega\}$ is infinite, closed and discrete (this uses the $T_1$-ness), so that $X$ has a non-compact countable closed set.