Proposition 6.5.3.label Let $X$ be a uniform space and $A \subset X$ be closed, then $A$ is complete.
Proof. Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of Definition 5.5.2.$\square$