Is there a theorem that says:
If $H$ is a Hilbert space and $U$ is any subspace then $$ H = U \oplus U^\bot$$ if and only if $U$ is closed?
My conjecture is yes.
I can easily prove that if $U$ is closed then $H = U \oplus U^\bot$. But it's not so clear whether if $H = U \oplus U^\bot$ then $U$ is closed. I mean, it seems to me that it should be true but it's not clear how I can show it. So maybe it's not even true.
Argue as follows: Since any $v\in H$ can be written $v=u+u^\perp$ and $\| v\|^2 =\| u\|^2 + \| u^\perp\|^2$, we get that the mapping $v\mapsto u$ is a continuous projection with range $U$. The result follows from general considerations:
If $T$ is a bounded projection (i.e. $T=T^2$), then the range of $T$ is closed. Indeed take $y_n=Tx_n$ a sequence in the range converging to some $y$. Then $$T(y)=\lim Ty_n=\lim T(Tx_n)=\lim Tx_n=y.$$ Therefore $y$ is in the range.