Subspace is compact iff it is closed

Question

Show that a metric subspace $(A,d_A)$ of $(E,d)$ is compact if, and only if, it is closed in $E$.

Is it true? It seems that only the direction $(\rightarrow )$ follows.

The proof of direction $(\rightarrow)$ is: Since $A$ is compact it is complete and totally bounded. But $(A,d_A)$ complete implies it is closed in $E$.

Answer 1

True if $(E,d)$ is a compact metric space. False otherwise: just take $A=E$.

Answer 2

This is not true in general.

E.g. $(\mathbb R,d)$ is a metric subspace of itself and is closed, but it is not compact.