 Solution

Let $(x_n)$ be any sequence in a complete, totally bounded metric space $X$. Cover $X$ by finitely many balls of radius $2^{-1}$; one contains infinitely many terms. From those terms choose an infinite subsequence lying in one ball of radius $2^{-2}$, and continue. The diagonal argument produces a subsequence $(x_{n_j})$ whose tail from its $j$th term onward lies in a ball of radius $2^{-j}$. Hencewhenever $k,\ell \ge j$, so the subsequence is Cauchy. Completeness makes it converge in $X$.

Every sequence in $X$ therefore has a convergent subsequence, so $X$ is sequentially compact. Sequential compactness is equivalent to compactness for metric spaces, and thus $X$ is compact. This proves that a complete totally bounded metric space is compact.