# Let X be sequentially compact. Then X is also countably compact.

## A subset $$A$$ of a topological space $$X$$ is sequentially compact iff every sequence in $$A$$ has a convergent subsequence.

Countably compact means every countable open cover has a finite subcover. I don't understand how can I link these two definitions to prove this.

Let $$X=\bigcup_n U_n$$ where $$U_n$$ is open for each $$n$$. Suppose there is no finite subcover. For each $$n$$ there exists $$x_n \in X\setminus \bigcup_{k=1}^{n} U_k$$. Let $$x$$ be the limit of some subsequence $$\{x_n'\}$$ of $$\{x_n\}$$. Then $$x \in U_j$$ for some $$j$$. Hence $$x_n' \in U_j$$ eventually but this is a contradiction to the choice of $$\{x_n\}$$.