= Solution
A metric space is <totally bounded> if, for every $\epsilon>0$, finitely many open balls of radius $\epsilon$ cover it. The Bolzano-Weierstrass property here means that every sequence has a subsequence converging to a point of the space, namely <sequential compactness>.
First suppose that property holds. A <Cauchy sequence> has a convergent subsequence with limit $x$ in the space. Given $\epsilon>0$, choose an index after which all sequence terms lie within $\epsilon/2$ of each other, and a subsequence term beyond that index within $\epsilon/2$ of $x$. The <triangle inequality> then puts every subsequent term within $\epsilon$ of $x$. Thus the whole Cauchy sequence converges, proving completeness.
If total boundedness failed, some $\epsilon>0$ would admit no finite ball cover. Choose points inductively outside the union of the $\epsilon$-balls about previously chosen points. Distinct sequence terms then have distance at least $\epsilon$. No subsequence can be Cauchy, and therefore none can converge, a contradiction. This proves total boundedness.
Conversely, suppose the space is complete and totally bounded. For any sequence, cover the space by finitely many radius-$2^{-1}$ balls and retain the infinitely many indices in one ball. Within those indices, a finite cover by radius-$2^{-2}$ balls retains an infinite subset in one smaller ball. Repeat with radii $2^{-k}$ to obtain nested infinite index sets $I_k$. Choose increasing indices $n_k\in I_k$. For $l,j\ge k$, both terms lie in the ball selected at step $k$, so
$$
d(x_{n_l},x_{n_j})<2^{1-k}.
$$
The subsequence is Cauchy and completeness makes it converge within the space. Hence \b[the Bolzano-Weierstrass property is equivalent to completeness plus total boundedness].
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