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 \overset{\sim}{X}[\itex] that are intersection of nullspaces of one dimensional subspaces of 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.