Sequentially Compact Metric Space is Compact/Proof 1

Theorem
A sequentially compact metric space is compact.

Proof
Let $\left({A, d}\right)$ be a sequentially compact metric space. Our proof is based on Sequentially Compact Metric Space is Lindelöf, that is, every open cover of $A$ has a countable subcover.

Take any open cover $C$ of $A$, and extract from it a countable subcover $\left\{{U_1, U_2, \ldots}\right\}$.

Reasoning by contradiction, assume that there is no finite subcover of $C$. Then, for any natural $n \ge 1$, the family $\left\{{U_1, \ldots, U_n}\right\}$ does not cover $A$, so we can choose a point $x_n \in A$ such that
 * $x_n \notin U_1 \cup \cdots \cup U_n$.

In this way we have constructed an infinite sequence $\left\{{x_n}\right\}_{n \ge 1}$ of points of $A$.

As we are assuming $A$ is sequentially compact, this sequence has a subsequence which converges to some $x \in A$. But there is some $U_m$ such that $x \in U_m$ (as the $U_i, i \ge 1$, are a cover), and hence by one of the characterizations of convergence, there is an infinite of number of terms in the sequence $\left\{{x_i}\right\}$ which are contained in $U_m$.

This is a contradiction, as in the way we constructed our sequence, each $U_n$ can only contain a finite number of the terms ($U_n$ can contain only points $x_i$ with $i < n$).