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**Aryth** Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

Suppose that $\displaystyle A \subseteq X$ is countably compact and let $\displaystyle B \subset A$ be closed. We will let $\displaystyle \mathcal{F} = \{F_i : i \in \mathbb{N}\}$ be a countable cover of $\displaystyle A$. Since $\displaystyle X \backslash B$ is open, it follows that $\displaystyle \mathcal{F}$ along with $\displaystyle X \backslash B$ is a countable open cover of $\displaystyle A$. We know that $\displaystyle A$ is countably compact, so $\displaystyle A$ can be covered by a finite number of sets from $\displaystyle \mathcal{F}$, $\displaystyle F_1, F_2, \cdots , F_n$, and possibly $\displaystyle X \backslash B$. Since $\displaystyle B \subset A$, $\displaystyle F_1, F_2, \cdots, F_n$ covers $\displaystyle B$. Therefore, $\displaystyle B$ is countably compact.