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

Question

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.

Answer 1

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 $\{x_n\}$.