= Solution
A <complete metric space> is a <metric space> in which every <Cauchy sequence> converges to a point of that space.
If $x_n\to x$, choose $N$ so that $d(x_n,x)<\varepsilon/2$ for all $n\geq N$. The <triangle inequality> gives $d(x_m,x_n)\leq d(x_m,x)+d(x,x_n)<\varepsilon$, proving the <Cauchy sequence> property.
Now let $A$ be complete in the induced <metric>. If $x\in\overline A$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a <Cauchy sequence>. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a <metric space> gives $a=x$, hence $\overline A\subseteq A$. Therefore \b[a complete <metric> subspace is closed].
Completeness does not imply <sequential compactness>. In the complete space $\mathbb R$, the sequence $x_n=n$ has no convergent <subsequence>: a <subsequence> has strictly increasing indices $n_k\geq k$, so $x_{n_k}\to+\infty$ and is unbounded, whereas every convergent real sequence is bounded.
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