I hope someone would be able to help me with the finer details of this proof.
Problem: $M$ is a non-empty set in an Inner Product Space (IPS) $X$. I need to show that the annihilator of $M$ which is defined by $M^\ast = \{x \in X \mid \langle x,y\rangle = 0\text{ for all } y \in M\}$ is a closed subspace of $X$.
My Question: I can do the proof if I assume $M$ is a closed subspace of a Hilbert space $H$, however this is not given above and I doubt I can just assume this.
Workings: I want to use the fact that $M^\ast$ is the Null Space of $P$ where $P$ is the orthogonal projection of $H$ onto $M$ if $M$ is closed and $H$ is a Hilbert space. And the Null Space is closed for a bounded linear operator $P$.
I know that for any IPS $X$ there exist a Hilbert space $H$ and an isomorphism $A: X \twoheadrightarrow Y$ where $Y$ is a dense subspace in $H$. I am not sure how to use this to derive that $M$ is closed.
Can I set "$A=P$" and "$Y=M$"? Then $M$ is countable since $H$ is separable? Then I can use: $M$ countable implies $M$ finite implies $M$ complete implies $M$ closed. And then the rest of my proof.
Hope this is clear on what I want to do.
For every (fixed for now) $y \in M$ we define a function $f_y: X \rightarrow \mathbb{R}$ by $f_y(x) = (x,y)$, (the inner product of $x$ and $y$). This is a continuous function (by Cauchy-Schwarz, e.g.) and so ${f_y}^{-1}[\{0\}]$ is closed in $X$, as the inverse image of a closed set in $\mathbb{R}$ under $f_y$.
Now note that $M^{\ast} = \bigcap_{y \in M} {f_y}^{-1}[\{0\}]$, which is an intersection of closed sets, so closed as well. No assumptions on $M$ are needed.