 Solution

First, compactness makes $X$ totally bounded: for every positive integer $k$, the open cover by balls of radius $1/k$ has a finite subcover. Given a sequence, repeatedly choose a ball of radius $1/k$ containing infinitely many terms of the subsequence retained at the preceding stage. Taking a diagonal subsequence $(x_{n_k})$, any two terms with indices at least $k$ lie in one ball of radius $1/k$, so their distance is below $2/k$. The diagonal subsequence is therefore a Cauchy sequence.

A compact metric space is complete. To see this without assuming sequential compactness, let $(y_n)$ be Cauchy and put $F_n=\overline{\{y_m:m\ge n\}}$. These nonempty closed sets are nested and have the finite intersection property; compactness gives a point $y\in {\bigcap }_n{F}_n$. Given $\epsilon >0$, a sufficiently late tail has diameter below $\epsilon /2$, and because $y$ lies in its closure, every point of that tail lies within $\epsilon$ of $y$. Thus $y_n\to y$.

The Cauchy subsequence constructed above consequently converges in $X$. Hence every compact metric space is sequentially compact.