How to show that $(W^\bot)^\bot=W$ (in a finite dimensional vector space)

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?

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?

$\dim W + \dim W^\perp =n$ implies $\dim W^\perp + \dim W^{\perp\perp} =n$ and so $\dim W = \dim W^{\perp\perp}$.

Now you just need to prove that $W \subseteq W^{\perp\perp}$ or $W \supseteq W^{\perp\perp}$, whichever is easier.

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