Suppose $(A,\tau)$ is a topological vector space and $B$ is a subset of $A$. Suppose $B$ is metrizable and complete (in this metric). Must $B$ be closed in $\tau$?
Here is my attempt:
First, I will add the assumption that the metric $d$ on $B$ is translation-invariant. Let $\tau_B$ be the induced subspace topology on $B$, so that the metric-topology generated by $d$ on $B$ is equal to $\tau_B$.
Since $d$ is translation invariant, these notes I found here imply that, since $(B,d)$ is complete, so is $(B,\tau_B)$ (in the sense that every Cauchy net in $(B,\tau_B)$ converges to a point in $B$).
Suppose $x$ belongs to the closure of $B$. Then there exists a net $x_\alpha$ of elements from $B$ converging to $x$. I was able to show that any convergent net in a topological vector space is a Cauchy net. Thus $x\in B$ by the completeness of $(B,\tau_B)$, which shows that $B$ is closed.
Remarks: First, I'm not sure if this reasoning is correct, which I am most concerned about. Secondly, I'm not sure what assumptions are necessary, particularly whether the metric must be translation-invariant, and whether this holds more generally in any topological space (not necessarily a topological "vector" space).
The simple answer is: no. Consider $A=\mathbb{R}$, the topology $\tau=\{\emptyset,A\}$ is anti-discrete and let $B=\{1\}$.
The problem with your reasoning is that $(x_\alpha)$ may be a Cauchy net in $B$ but in $A$ that doesn't even make sense (in order to talk about Cauchy nets we need metric or uniform structure). And even if $A$ was metric the concepts of Cauchy nets in $A$ and $B$ may differ, since metric in $A$ and $B$ are very loosely related.