Given a topological space $A$, we can endow a subset $B\subseteq A$ with the subspace topology.
We say that a subset $C\subseteq B$ is "open relative to $B\,$" when $C$ is open in the subspace topology on $B$; that is, by definition, when there is some open subset $U$ of $A$ such that $C=U\cap B$.
The definition is identical for "closed", i.e. a subset $C\subseteq B$ is "closed relative to $B\,$" when it is closed in the subspace topology on $B$, which (by definition) is when there is a closed subset $D$ of $A$ such that $C=D\cap B$.
Thus, it is a trivial result that every topological space $A$ is closed relative to $A$, because $A$ must be closed in its topology (that is part of the definition of a topology) and therefore $A=A\cap A$ is the intersection of a closed subset of $A$ with $A$.
Any metric space $X$, with distance function $d$, can be given a topology where the open subsets of $X$ are precisely those subsets $Y\subseteq X$ with the property that, for any $p\in Y$, there is an $\epsilon>0$ such that $B_\epsilon(p)\subseteq Y$, where $B_\epsilon(p)=\{q\in X\mid d(p,q)<\epsilon\}$ is the open ball of radius $\epsilon$ centered at $p$. Thus, any metric space can be given the structure of a topological space.
Thus, we can simply apply the above result (which holds for all topological spaces) to a metric space, and obtain the statement that any metric space $X$ is closed relative to $X$.