Sum of subspaces is closed => the subspaces were closed? I have the following question:
We know that if $A$ is a closed subspace of a Hilbert space $\mathcal{H}$ and $B$ is a finite dimensional subspace such that $A\cap B=\{0\}$, then $A\dot{+}B$ is necessarily closed. Does this also go the other way round, i.e. if we take the direct sum of two subspaces $A$ and $B$ with $A\cap B=\{0\}$, where $B$ is finite dimensional and we are given that $A\dot{+} B$ is closed, can we conclude that $A$ is closed?

No. A one dimensional subspace is always closed, but a subspace of codimension one need not be, if the space is infinite dimensional.
Indeed, if $H$ is an infinite dimensional Hilbert space, there exists a non continuous (that is, unbounded) linear form $f\colon H\to \mathbb{R}$ (or $\mathbb{C}$, depending whether you're dealing with real or complex vector spaces).
Take a complement $B$ of $\ker f$. Then $\ker f\cap B=\{0\}$ by definition, $\ker f+B=H$ (hence closed), $B$ is closed (finite dimensional, hence complete, hence closed), but $\ker f$ is not closed.
The standard example is $l^2$, which is not isomorphic (as vector spaces) to its algebraic dual, while being isomorphic to its topological dual, so it admits non continuous linear forms.
Of course this counterexample works in the standard setting where the axiom of choice is assumed. Asaf Karagila has shown in a comment that, under different assumptions, the algebraic dual of $l^2$ can be the same as the topological dual.
