If $H$ is a Hilbert space and $U_1$ and $U_2$ are closed orthogonal subspaces, then $U_1 \oplus U_2$ is a closed I have a question asking for a proof to show that if $H$ is a Hilbert space and $U_1$ and $U_2$ are closed orthogonal subspaces, then $U_1 \oplus U_2$ is  closed.
The proof starts of by saying 
We prove that $$\overline {U_1 \oplus U_2} \subset U_1 \oplus U_2$$ 
How does $\overline {U_1 \oplus U_2} \subset U_1 \oplus U_2$ relate to the question, what is it saying?
Why is the closure of $U_1 \oplus U_2$ a subset of $U_1 \oplus U_2$?
 A: The proof starts that way because that's the only part of the proof that's non-trivial. A set $A$ is closed iff $A = \overline{A}$. We always know that $A \subseteq \overline{A}$. So the only thing left to prove that a set $A$ is closed, is the other inclusion $\overline{A} \subseteq A$.
Here it is applied to $A = U_1 \oplus U_2$
A: This is Surb's answer detailed:
Let $\{x_n\}\subset  U_1\oplus U_2$ be a convergent sequence. Let denote $\ell$ it's limit. Then, for all $n$, there are unique $x_n^1\in U_1$ and $x_n^2$ s.t. $x_n=x_n^1+x_n^2$ with $\left<x_n^1,x_n^2\right>=0$.
Let $limsup_{n,m}=lim_{N \to \infty} sup_{n,m>N}$.
Now $limsup_{n,m} |x^1_n-x^1_{m}|^2 \leq limsup (|x^1_n-x^1_{m}|^2+|x^2_n-x^2_{m}|^2)=lim_{N \to \infty} sup_{n,m>N}|x_n-x_{m}|^2=0$.  Thus  the limit of $x_n^1$, denoted $\ell_1$ exists and by hypothesis is in $U_1$.  Similarly $\ell_2$, the limit of $x^2_n$ exists and is in $U_2$. 
By sum rule of limit, $\ell_1 +\ell_2=\ell$.  Since $\ell_1, \in U_1$ and $\ell_2 \in U_2$, $\ell \in U_1 \oplus U_2$, so that the limit of $x_n$ is in $U_1 \oplus U_2$. 
Nothing else Surb asked for needs to be shown.
