Sequentially Compact Metric Space is Compact/Proof 1

Theorem

Let $C$ be an open cover $A$.

Extract from it a countable subcover $\set {U_1, U_2, \ldots}$.

Aiming for a contradiction, suppose there exists no finite subcover of $C$.

Then, for all $n \in \N_{\ge 1}$, the set $\set {U_1, \ldots, U_n}$ does not cover $A$.

Hence it is possible to choose $x_n \in A$ such that:

$x_n \notin U_1 \cup \cdots \cup U_n$.

Thus we construct an infinite sequence $\sequence {x_n}_{n \mathop \ge 1}$ of points of $A$.

By assumption $A$ is sequentially compact.

Thus $\sequence {x_n}_{n \mathop \ge 1}$ has a subsequence which converges to some $x \in A$.

But because the $U_i: i \ge 1$ forms a cover for $A$, there exists some $U_m$ such that $x \in U_m$.

Hence by Limit of Sequence is Accumulation Point, there is an infinite number of terms in the sequence $\sequence {x_i}$ which are contained in $U_m$.

But from the method of construction of $\sequence {x_i}$, each $U_n$ can contain only points $x_i$ with $i < n$.

That is, each $U_n$ can only contain a finite number of the terms of $\sequence {x_i}$.

Thus the supposition that there exists no finite subcover of $C$ was false.

Hence the result.

$\blacksquare$