Sequentially Compact Metric Space is Compact/Proof 1

Theorem
A sequentially compact metric space is compact.

Proof
Let $(X,d)$ be a sequentially compact metric space. Our proof is based on the result that a sequentially compact metric space is Lindelöf, that is, every open cover of $X$ has a countable subcover.

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

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

In this way we have constructed an infinite sequence $\{x_n\}_{n \geq 1}$ of points of $X$.

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