Complete subspaces are closed

Statement

Let $(X, d)$ be a metric space and let $A \subseteq X$. If $A$ with the subspace metric is complete, then $A$ is closed in $X$.

Remarks

Proof: let $x \in \overline{A}$. By closure is the set of limits of sequences, there is a sequence in $A$ with limit $x$. It is a Cauchy sequence in $A$, so it has a limit in $A$. Limits are unique in $X$, so $x \in A$. The space $X$ need not be complete here. For example, $\mathbb{Q}$ is not closed in $\mathbb{R}$, so $\mathbb{Q}$ is not complete. In a complete space, a subset is complete if and only if it is closed, by closed subsets of complete spaces are complete.

Explanations

L1 human

A complete subset contains the limits of its Cauchy sequences, and thus of its convergent sequences. A set that keeps all its limits is closed.