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: 
You have a finite-dimensional vector space $V$ of dimension $n$, and you have a subspace $W\subseteq V$.

Notice first that $W \subseteq W^{\perp\perp}$:
If $x \in W$, then $\langle x, u \rangle = 0$ for all $u \in W^{\perp}$.
Since an element $y\in W^{\perp\perp}$ is defined by the property that $\langle y, u \rangle = 0$ forall $u \in W^{\perp}$, you see that $x \in W^{\perp\perp}$.

From $\dim W + \dim W^{\perp} = n$ and $\dim W^{\perp} + \dim W^{\perp\perp}=n$, you get

$$
\dim W +\dim W^{\perp}
= \dim W^{\perp} + \dim W^{\perp\perp},
$$

or

$$
\dim W 
= \dim W^{\perp\perp}.
$$

Since $W\subseteq W^{\perp\perp}$, you have $W=W^{\perp\perp}$.
**Hint** It follows from the definition that $(W^\perp)^\perp \subset W$.

**Hint 2:** For every subspace $U$ of $V$ you have
$$\dim(U)+ \dim(U^\perp)=\dim(V)$$

What does this tells you about $\dim(W)$ and $\dim (W^\perp)^\perp$?