Decomposition of a Hilbert Space

Question

I'm trying to show the following:

In the decomposition $x=a+b$ with $a\in M$ and $b\in M^{\perp}$, $a$ and $b$ are unique. Deduce that if $H=M\oplus N$, where $M$ is a closed linear subspace and $M\perp N$, then $N=M^{\perp}$.

I feel like I may be simplifying things too much, so I would greatly appreciate feedback.

If $y\in N$, then $\langle y,x\rangle=0$ for all $x\in M$. Thus, $y\in M^{\perp}$ by definition, so $N\subseteq M^{\perp}$.

Next, let $x\in M^{\perp}$. Since $H=M\oplus N$ where $M\perp N$, we have $x=a+b$ where $a\in M$ and $b\in N$. But since $N\subseteq M^{\perp}$ we have $a=0$, thus $x=b\in N$, giving us $M^{\perp}\subseteq N$, and hence $M^{\perp}=N$.

Answer 1

Looks good except for one part.

But since $N\subseteq M^{\perp}$ we have $a=0$

This step could use some more explanation. Why does $N\subseteq M^\perp$ imply $a=0$? How exactly are you using the uniqueness of the decomposition of $x$ as stated in the problem (or do you have some different argument in mind)?