= Solution
Let $(x_n)$ be a <Cauchy sequence> in a compact metric space $X$. By <sequential compactness of a compact metric space>, it has a <convergent subsequence> $x_{n_j}\to x\in X$. Given $\varepsilon>0$, choose $N$ such that $d(x_m,x_n)<\varepsilon/2$ whenever $m,n\geq N$, and then choose $j$ such that $n_j\geq N$ and $d(x_{n_j},x)<\varepsilon/2$. For every $n\geq N$, the <triangle inequality> gives
$$
d(x_n,x)\leq d(x_n,x_{n_j})+d(x_{n_j},x)<\varepsilon.
$$
\b[Thus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a <complete metric space>.]
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