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A G-delta set is a countable intersection of open subsets of a topological space. In a complete metric space these are precisely the topologically complete subspaces, by the G-delta criterion for topological completeness.

A subspace is topologically complete if its subspace topology is induced by some complete metric. The compatible complete metric need not be the restriction of the displayed ambient metric. For example, an open interval is topologically complete although its usual metric is incomplete.

Suppose $\rho$ is a compatible complete metric on $Y$. For each $n\ge 1$ and $y\in Y$, choose an ambient open set $V_{n,y}$ containing $y$, contained in $B_d(y,1/n)$, and with $\rho$-diameter of $V_{n,y}\cap Y$ at most $1/n$. Such sets exist because the two topologies on $Y$ agree. Put $U_n={\bigcup }_{y\in Y}{V}_{n,y}$. Clearly $Y\subseteq {\bigcap }_n{U}_n$.

If $x\in {\bigcap }_n{U}_n$, choose $y_n\in Y$ with $x\in V_{n,y_n}$. Then $d(x,y_n)<1/n$, so $y_n\to x$ in the ambient metric. For fixed $m$, the point $x$ belongs to some open $V_{m,z}$, and eventually all $y_n$ belong to this set. Their pairwise $\rho$-distances are then at most $1/m$. Hence $(y_n)$ is $\rho$-Cauchy, and completeness gives a limit $y\in Y$. Compatibility gives $d(y_n,y)\to 0$, forcing $x=y$. ThusThis proves the G-delta criterion for topological completeness in the required direction.

The converse is true in a complete metric ambient space. If $Y={\bigcap }_n{U}_n$ with each $U_n$ open in complete $X$, define $a_n(y)=1/d(y,X\setminus U_n)$, omitting any term whose complement is empty. A compatible complete metric on $Y$ isContinuity of each $a_n$ and the uniformly small series tail show that $D$ has the original topology. A $D$-Cauchy sequence is $d$-Cauchy, hence converges to some $x\in X$. For each $n$, its real coordinates $a_n$ form a Cauchy sequence and remain bounded. Since distance to a closed set is continuous, $d(x,X\setminus U_n)>0$, so $x\in U_n$ for every $n$. The same coordinate convergence and series-tail argument give convergence in $D$. Thus $D$ is complete.

For the normed-space claim, embed $E$ densely in its norm completion $\hat{E}$. Topological completeness and the preceding criterion make $E$ a dense G-delta set in $\hat{E}$, hence comeagre. Every translate $x+E$ is also comeagre. The Baire category theorem in the complete space $\hat{E}$ implies that $E\cap (x+E)$ is nonempty. If $z=x+e$ lies in that intersection, then $x=z-e\in E$. This holds for every $x\in \hat{E}$, soThis is the comeagre subgroup completeness argument.

Finally let $B=\{\ell \in E^{\ast }:\Vert \ell \Vert <1\}$ with its weak-star topology. Put $r_n=1-1/(n+1)$ and $F_n=\{\ell \in B:\Vert \ell \Vert \le r_n\}$. The dual norm is the supremum of the continuous evaluation moduli on the unit ball of $E$, so every $F_n$ is relatively closed. Also $B={\bigcup }_n{F}_n$.

Each $F_n$ has empty relative interior. Given $\ell \in F_n$ and a basic weak-star neighborhood restricting finitely many evaluations on $x_1,\dots ,x_k$, the Hahn-Banach theorem gives a nonzero functional $h$ annihilating their finite-dimensional span, because $E$ is infinite-dimensional. All $\ell +th$ retain those evaluations. Their norm varies continuously with $t$ and becomes unbounded; choose $t$ so that $r_n<\Vert \ell +th\Vert <1$. This point lies in the neighborhood inside $B$ but outside $F_n$. Thus $B$ is a nonempty countable union of relatively closed nowhere dense sets and is not a Baire space. ConsequentlyThis weak-star open dual ball category obstruction works without assuming that $E$ is separable or that the weak-star ball is metrizable.