Let $\mathcal{H}$ be an infinite-dimensional, complex Hilbert space, and let $V,W$ be infinite-dimensional vector subspaces of $\mathcal{H}$ such that $V\cap W=\{\mathbf{0}\}$, and such that $$ \mathcal{H}=\overline{V\oplus W}, $$ where $\overline{\;\cdot\;}$ denote the closure operation. I would like to know under what conditions, if any, it is true that $$ \mathcal{H}=\overline{V}\,\oplus\,\overline{W}\,. $$ If it can help, we may assume that $$ \langle v|w\rangle=0 $$ for all $v\in V$ and for all $w\in W$.
Attempt to solution
First of all, let me note that every $x\in\mathcal{H}$ may be written as the limit of some sequence $\{x_{n}\}_{n\in\mathbb{N}}$ where $$ x_{n}= v_{n} + w_{n} $$ with $v_{n}\in V$ and $w_{n}\in W$ for all $n\in\mathbb{N}$. Furthermore, the fact that $\langle v,w\rangle=0$ for all $v\in V$ and for all $w\in W$ implies that $\{x_{n}\}_{n\in\mathbb{N}}$ is a Cauchy sequence if and only if $\{v_{n}\}_{n\in\mathbb{N}}$ and $\{w_{n}\}_{n\in\mathbb{N}}$ are both Cauchy sequences on their own. Indeed, we have $$ ||x_{n} - x_{m}||^{2}=\langle x_{n}-x_{m}|x_{n}-x_{m}\rangle = \\ =\langle (v_{n}-v_{m}) + (w_{n} - w_{m})|(v_{n}-v_{m}) + (w_{n} - w_{m})\rangle = \\ = \langle v_{n}-v_{m}|v_{n}-v_{m}\rangle + \langle w_{n} - w_{m}|w_{n} - w_{m}\rangle=\\=||v_{n}-v_{m}||^{2} + ||w_{n}-w_{m}||^{2} \,. $$ Clearly, the limit of $\{v_{n}\}_{n\in\mathbb{N}}$ is in $\overline{V}$, and the limit of $\{w_{n}\}_{n\in\mathbb{N}}$ is in $\overline{W}$ (by definition).
From this, we have that every $x\in\mathcal{H}$ may be written as the limit of the sum of two Cauchy sequences. Since, for Cauchy sequences, the limit of the sum is equal to the sum of the limits, we obtain that every $x\in \mathcal{H}$ may be written as $$ x=\overline{v} + \overline{w} $$ with $\overline{v}\in\overline{V}$ and $\overline{w}\in\overline{W}$.
All that is left to check is that $\overline{V}\cap\overline{W}=\{\mathbf{0}\}$. At this purpose, we first suppose that $\overline{V}\cap\overline{W}\neq\{\mathbf{0}\}$, that is, there exists a nonzero vector $x$ which is both in $\overline{V}$ and in $\overline{W}$, and then obtain a contradiction. Then, since $x\in\overline{V}$, there is a Cauchy sequence $\{v_{n}\}_{n\in\mathbb{N}}$, with $v_{n}\in V$ for all $n\in\mathbb{N}$, such that $x$ is the limit of $\{v_{n}\}_{n\in\mathbb{N}}$. Now, we pick an arbitrary element $w\in W$ so that we have $$ 0=\langle w|x\rangle=\lim_{n}\,\langle w|v_{n}\rangle=0 \,. $$ Since $w$ is arbitrary in $W$ and $W$ is dense in $\overline{W}$, we conclude that $x=\mathbf{0}$, and we reach a contradiction.