Sum of closed subspaces of a Hilbert space is closed Let $M, N ⊂ H$ ($H$ Hilbert), be two closed linear subspaces.
Assume that $\langle u, v\rangle = 0$ $∀u ∈ M$, $∀v ∈ N$. Prove that $M + N$ is closed.
Take a sequence $(g_n)\in M+N$ such that $g_n\to x\in H$.
Then for any $n\geq 1$, $\exists v_n\in N, u_n\in M$ such that $g_n=v_n+u_n$. This implies that the sequences $(v_n)$ and $(u_n)$ converge singularly to the elements $v,u\in H$, and by closeness $v\in N$ and $u\in M$.
By uniqueness of the limit it must hold $x=u+v$ which implies $x\in N+M$.
I did not use any property of orthogonality so I guess this reasoning is wrong. Why? And how should I use orthogonality?
 A: Without orthogonality this is false: an example is given by Robert Israel.
Orthogonality implies that $\|u+v\|^2 = \|u\|^2+\|v\|^2$ for $u\in M$, $v\in N$. Thus, if a sequence $(u_n+v_n)$ converges, the inequalities such as
$$\|u_n-u_m\|\le \|(u_n+v_n)-(u_m+v_m)\|$$
imply convergence of both $u_n$ and $v_n$. So, $u_n\to u\in M$ and $v_n\to v\in N$, which implies $\lim(u_n+v_n  ) = u+v\in M+N$.
A: Let's do a proof without using limits.
The following fact can be found in any book of Functional Analysis, its proof is easy:

Fact 1: Let $G$ be a subspace of a Hilbert space $H$, then $G^{\bot \bot}=\overline{G}$.

Therefore, we have to show that $(M+N)^{\bot\bot}\subseteq M+N$. In fact, let $x\in (M+N)^{\bot\bot}$, then 
$$\left\langle x,y \right\rangle =0 \qquad \forall y\in (M+N)^{\bot}. \tag{I}$$
Another fact that is easy to show is the following:

Fact 2: $(M+N)^{\bot}=M^{\bot}\cap N^{\bot}.$

Since $M$ is closed then $H=M\oplus M^{\bot}$, then there exists $m\in M$ and $m^{\bot}\in M^{\bot}$ such that $x=m+m^{\bot}$.
But we know that $M\bot N$ then $N\subset M^{\bot}$, then we have to show that $m^{\bot} \in N$. 
In fact, note that if $m^{\bot}=0$ then we finished, there would be nothing to show. So,  we suppose that $m^{\bot} \notin N$ with $m^{\bot}\neq 0$, then since $N$ is closed we have $H=N\oplus N^{\bot}$, so $m^{\bot}\in N^{\bot}$. Therefore, $m^{\bot}\in M^{\bot}\cap N^{\bot}$, then by Fact 2 we have that $m^{\bot}\in (M+N)^{\bot}$. Therefore, by Fact 1 we have 
$$0=\left\langle x,m^{\bot} \right\rangle =\left\langle m+m^{\bot},m^{\bot} \right\rangle=\left\langle m^{\bot},m^{\bot} \right\rangle=\left\|m^{ \bot}\right\|^{2}.$$
Then $m^{\bot}=0$, what contradicts $m^{\bot}\neq 0$.
