Suppose a Hilbert space $\mathcal{H}$ and $\mathcal{S}\subset\mathcal{H}$ a subspace. Now, $\mathcal{S}$ can be closed or open in the norm topology depending on whether or not every convergent sequence of elements of $\mathcal{S}$ has a limit in $\mathcal{S}$. If $\mathcal{S}$ is closed, then it is also a complete space, as every convergent sequence is also Cauchy convergent. Suppose then that $\mathcal{S}$ is complete, it is then necessarily closed?
Any counterexample?
Given a sequence $(x_{n})\subseteq S$ such that $x_{n}\rightarrow x$ in $H$, then $(x_{n})$ is Cauchy in $S$. As $S$ is complete, then $x_{n}\rightarrow y$ in $S$, that is, for some $y\in S$. So $x_{n}\rightarrow y$ in $H$. As $H$ is Hausdorff, then $x=y$, so $S$ is closed.