Sequentially Compact Metric Space is Complete

Theorem
Let $M = \left({A, d}\right)$ be a metric space which is sequentially compact.

Then $M$ is complete.

Proof
Let $\left \langle {x_n}\right \rangle_{n \in \N}$ be a Cauchy sequence in $A$.

As $M$ is sequentially compact, $\left \langle {x_n}\right \rangle$ has a convergent subsequence.

By Convergent Subsequence of Cauchy Sequence in Metric Space, this implies that the entire sequence $\left \langle {x_n}\right \rangle$ is convergent.

Hence, $M$ is complete.