Let $X$ be a metric space, and let $Y$ be a complete subspace of $X$. Then $Y$ is closed.

Proof

Let $x\in\overline{Y}$ be a point in the closure of $Y$. Then by the definition of closure, from each ball $B(x,\frac{1}{n})$ centered in $x$, we can select a point $y_{n}\in Y$. This is clearly a Cauchy sequence in $Y$, and its limit is $x$, hence by the completeness of $Y$, $x\in Y$ and thus $Y=\overline{Y}$.