Proposition 8

Every closed subspace of a complete space is complete. Every complete subspace of a Hausdorff uniform space (complete or not) is closed.

Let $X$ be a complete space and let $A$ be a closed subspace of $X$. If $\mathfrak{F}$ is a Cauchy filter on $A$, then it is a Cauchy filter base on $X$ (no. 1, Proposition 3) and therefore converges to a point $x \in X$; but since $A$ is closed we have $x \in A$, and therefore $\mathfrak{F}$ converges in the subspace $A$.

Now let $\overline{A}$ be a non-closed subset of a Hausdorff uniform space $X$, and let $b \in \overline{A} - A$. The trace $\mathfrak{B}_A$ on $A$ of the neighbourhood filter $\mathfrak{B}$ of $b$ in $X$ is a Cauchy filter on $A$; but it cannot converge to a point $c \in A$, otherwise $c$ would be a limit point of $\mathfrak{B}$ (no. 2, Proposition 5, Corollary 3) which is absurd since $b \neq c$ and $X$ is Hausdorff.