Statetment
Let $M \subset H$ be nonempty, and $H$ is a Hilbert space. Then $M^\perp$ is a subspace, and $M^\perp $ is closed.
Proof
To show that $M^\perp$ is a subspace we let $x,y \in M^\perp$, $a,b \in \mathbb{R}$, and let $v \in M$ be arbitrary. Then
$$\langle ax + by , v \rangle = a\langle x, v \rangle + b \langle y, v \rangle = 0$$ which implies $ax + by \in M^\perp$.
Now, to show $M^\perp$ is a closed subspace of $X$, we must show that if $x$ is a limit point of $M^\perp$ that $x$ is in $M^\perp$. So, let $x$ be a limit point of $M^\perp$. By definition of a limit point, there exists a sequence $(x_n)$ in $M^\perp$ such that $x_n \longrightarrow x$ as $n \longrightarrow \infty$. And, since $x_n \in M^\perp$, we have $\langle x_n, v \rangle = 0$ for all $n = 1, 2, ...$ and $v \in M$. So, by continuity of the inner product we observe
\begin{equation} 0 = \langle x_n, v \rangle \tag{$v\in M$}\\ \text{ } \\ \implies 0 = \lim_{n\to \infty} \langle x_n, v \rangle = \langle \lim_{n\to \infty} x_n ,v \rangle = \langle x, v \rangle \end{equation} where $v \in M$ is arbitrary. So, $x \perp M$, and therefore $x \in M^\perp$. And since $x$ is an arbitrary limit point of $M^\perp$, we conclude that $M^\perp$ is closed.
Just looking for verification and/or comments on clarity. Thank you.