= Solution
A <Cauchy sequence> in a <metric space> satisfies: for every $\varepsilon>0$ there is $N$ such that $d(x_m,x_n)<\varepsilon$ whenever $m,n\geq N$. Suppose a <subsequence> $x_{n_k}$ converges to $x\in X$. Choose $N$ so that the Cauchy bound is $\varepsilon/2$, and then choose $k$ with $n_k\geq N$ and $d(x_{n_k},x)<\varepsilon/2$. For every $n\geq N$, the <triangle inequality> gives
$$
d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x)<\varepsilon.
$$
Thus \b[the whole Cauchy sequence converges to the subsequential limit].
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