= Solution
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 $n\ge1$ 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 $Y\subseteq\bigcap_nU_n$.
If $x\in\bigcap_nU_n$, 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 $d(y_n,y)\to0$, forcing $x=y$. Thus
$$
\boxed{Y=\bigcap_{n\ge1}U_n.}
$$
This proves the <G-delta criterion for topological completeness> in the required direction.
\b[The converse is true in a complete <metric> ambient space.] If $Y=\bigcap_nU_n$ 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 $Y$ is
$$
\boxed{D(y,z)=d(y,z)+\sum_{n\ge1}2^{-n}\min\{1,|a_n(y)-a_n(z)|\}.}
$$
Continuity 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 $\widehat E$. Topological completeness and the preceding criterion make $E$ a dense <G-delta set> in $\widehat E$, hence <comeagre>. Every translate $x+E$ is also comeagre. The <Baire category theorem> in the complete space $\widehat E$ 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 $x\in\widehat E$, so
$$
\boxed{E=\widehat E\text{ and }E\text{ is a Banach space}.}
$$
This is the <comeagre subgroup completeness argument>.
Finally let $B=\{\ell\in E^*: \|\ell\|<1\}$ with its <weak-star topology>. Put $r_n=1-1/(n+1)$ and $F_n=\{\ell\in B:\|\ell\|\le r_n\}$. 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 $B=\bigcup_nF_n$.
Each $F_n$ has empty relative interior. Given $\ell\in F_n$ and a basic weak-star neighborhood restricting finitely many evaluations on $x_1,\ldots,x_k$, 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 $r_n<\|\ell+th\|<1$. 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>. Consequently
$$
\boxed{B\text{ with its weak-star topology is not topologically complete}.}
$$
This <weak-star open dual ball category obstruction> works without assuming that $E$ is separable or that the weak-star ball is metrizable.
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