Double Orthogonal Closed Subspace Inner Product => Hilbert

[SqueeSpleen]

Let $X$ be an Inner Product Space. If for every closed subspace $M$, $M^{\perp \perp} = M$, then $X$ is a Hilbert Space (It's complete).
Hint: Use the following map: $T : X \longrightarrow \overset{\sim}{X}: T(y)=(x,y)=f(x)$ where $(x,y)$ is the inner product of $X$.

Relevant equations:
$S^{\perp}$ is always closed to every $S \subset X$

Attemp to solution.
I don't really know how to solve it, most theorems I have read, have Hilbert Space as a hyphotesis.
The only idea I had was trying to prove that $\overline{T(M^{\perp})} = M^{\circ}$ (Where $M^{\circ}$ is the Null Space of $M$).Edit: How do I do to have latex without jumping lines? I uploaded a pdf with a itext more closely formated to what I tried to write.

Edit2: I read other posts to find out how to do it, I only had to add an i to the "tex".

 

[SqueeSpleen]

I'm thinking about the family of subspaces of [itex]\overset{\sim}{X}[\itex] that are intersection of nullspaces of one dimensional subspaces of [itex]X[\itex], they are all closed. I think that if I can probe that they're all closed subspaces of $X[\itex] I may, somehow, be able to prove the required statement.$[/itex][/itex]