Question: I need to prove that if $V$ is a finite dimensional vector space over a field K with a non-degenerate inner-product and $W\subset V$ is a subspace of V, then:
$$
(W^\bot)^\bot=W
$$
Here is my approach:

If $\langle\cdot,\cdot\rangle$ is the non-degenerate inner product of $V$ and $B={w\_1, ... , w\_n}$ is a base of $V$ where ${w\_1, ... , w\_r}$ is a base of $W$ then I showed that 
$$
\langle u,v\rangle=[u]^T\_BA[v]\_B
$$
for a symmetric, invertible matrix $A\in\mathbb{R}^{n\times n}$. Then $W^\bot$ is the solution space of $A\_rx=0$ where $A\_r\in\mathbb{R}^{r\times n}$ is the matrix of the first $r$ lines of $A$. Is all this true?

I tried to exploit this but wasn't able to do so. How to proceed further?

Answer: 
It is clear that $W \subset W^{\perp \perp}$. Now, $\dim(W^{\perp \perp})=n-(n-\dim(W))=\dim(W)$. Now, when does a subspace have the same dimension as the space?
It is obvious that $W\subseteq W^{\perp\perp}$ suppose $\color{red}{x\in W^{\perp\perp}}$ and $\color{green}{x\notin W}$, since $H=\dim W \oplus \dim W^\perp$, so $\color{red}{x\in W^{\perp}}$ so $x=0 $ contradiction with $0\in W\cap W^\perp=\{0\}$