Double orthogonal complement of any closed subspace is it self Let $H$ be a pre-Hilbert space such that any closed sub space $M \subset H$ has the property $M^{\bot \bot}=M$. Prove that $H$ is a Hilbert space (ie, prove that $H$ is complete)

My attempt: 

As $H^*$ ($L(H,\mathbb{K})$) is complete then I want to prove that $H \cong H^*$. But I don't know how to construct an isometry between them. 

I will appreciate highly who can give me some ideas

Thank in advance!
That space seems to be the same space here:
A counterexample to theorem about orthogonal projection
and it doens´t seems to have the property $M = M^{\perp \perp}$ for all closed subspaces $M$.

I found the same exercise in the book "Functional Analysis" of George Bachman and Lawrence Narici

It says:
If X is an inner product space and if $M=M^{\perp \perp}$ for every closed subspace M of X, show that X is a Hilbert space. [Hint Use the mapping T mentioned in Eq. (12.23)]

That mapping is:

$T:X \rightarrow \overset{\sim}{X}$ such that

$y \rightarrow Ty = f $

where the bounded functional $f$ is, for any given $x \in X$, given by:

$(Ty)(x)=f(x)=(x,y)$

Edit:
I think I was able to prove it:

Let's be $T:X \to \overset{\sim}{X}$ such that
$T(y)=f_{y}$ where $f_{y}(x)=(x,y) \forall x \in X$

And $\overline{T}$ to the same function but with domain $\overline{X}$  (the completation of X).

It's clear that the conjugate space of $X$ and the conjugate space of $\overline{X}$ are the same, in fact the function $\overline{T}$ is an isometry, is aditive and homogeneuos conjugate (it's almost lineal and it's almost a congruency).

Let's suppose, to arrive a contradiction, that there exists $x \in \overline{X} - X$

It's clear that $x_{0} \neq 0$ given that $0 \in X$ because $X$ it's a vector space.

Let $f=\overline{T} (x_{0})$

We have that $N_{f}= \{ z \in X / f(z)=0 \}$ is a closed subspace of $X$, because it's the inverse image of $0$ under the bounded functional (so, a continous functional) $f$

As $x \neq 0$ we have that $f \neq 0$ so $N_{f} \neq X$

Let's see that $N_{f}^{\perp} = \{ 0 \}$ so we can arrive to a contradiction because $N_{f}=N_{f}^{\perp \perp} = \{ 0 \}^{\perp} = X$ (where the first equality is given by hyphotesis)

We have $ N_{f}^{\perp} = \{ x \in X / (y,x) = 0 \forall y \in X \}$

Then we have $x \in N_{f}^{\perp}$ which is the same than saying $(y,x) = 0 \forall y \in X$ then $T(x)(y)=0 \forall y \in X$ so $T(x)=\lambda f=\lambda T(x_{0})$

As $\left [ x_{0} \right ] \cap X = \{ 0 \}$ given that $x_{0} \notin X$, we arrive the desired contradiction.

Then $X=\overline{X}$, in other words $X$ is a Hilbert Space.

Sorry for the bad redaction, English isn't my first language.
This statement appears to be false:

Let $H=\ell^2(\mathbb R)$ be the Hilbert space with the usual orthonormal basis sequence $(e_n)_{n \in \mathbb N}$. Let $L$ denote the linear span of $e_n$.

Then $L$ is an (incomplete) inner product space: $$ s_n =\sum_{k=0}^n {1\over 2^k}e_k$$

converges to the sequence $s = (1,{1\over 2}, {1\over 4},\dots)$ which is not in $L$ but $s_n \in L$.

Since $e_n$ are orthonormal it is easy to see that if $M$ is any subspace of $L$ then $M^{\bot \bot}=M$.