 Solution

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence