Let $(E,d)$ be a metric space. If $S\subseteq E$ is closed in $E$, then $(S,d)$ is complete.
I was asked to prove this statement, but I feel that it is not correct. With the added statement of $(E,d)$ being complete, I can prove the statement as follows:
If $S$ is closed, then $S=\overline S$, where $\overline S$ is the closure of $S$. Hence, any convergent sequence of points in $S$ has limit in $S$. Consider a Cauchy sequence $(a_n)$ of points in $S$. This is a Cauchy sequence of points in $E$ since $S\subseteq E$, and since $E$ is complete, $(a_n)$ is convergent in $E$. Now since $S$ is closed and hence contains all limits of convergent sequences of points in $S$, $(a_n)$ converges in $S$. Since any Cauchy sequence of points in $S$ conveges in $S$, $S$ is Cauchy-complete.
This answers the question in the case of $(E,d)$ being complete unless I made some mistake. But in the problem statement we are not given that $(E,d)$ is complete. Can we still conclude that $(S,d)$ is complete after relaxing the condition that $(E,d)$ be complete?
You are right. If the whole space is not complete, then the whole space is an example of a closed space of itself which is not complete.