Let V be an inner product space, S and $S_0 $be subsets of V, and W be a finite-dimensional subspace of V. Let V be an inner product space, S and $S_0 $be subsets of V, and W be a finite-dimensional subspace of V. 
Prove:$S_0 \subset S$ implies $S^{\bot} \subset S_0^{\bot}$
Pf: Let $x \in S^{\bot}$, for all $y \in S$, the inner product $\langle x,y \rangle$ is given $\langle x,y \rangle=0$, which implies $y \in S_0$ Then $S_0 \subset S$, $x \in S_0^{\bot}$.Proved.
Prove: $W=(W^{\bot})^{\bot}$
Pf: Let $x \in W$ and $y \in W^{\bot}$. 
By definition, for all $y \in W^{\bot}$, the inner product is given by $\langle x,y \rangle=0$.
Using $x \in W$, this gives $W \subset (W^{\bot})^{\bot}$ How to prove the other side?
 A: Suppose that $w\in (W^\perp)^\perp$ then $\left<w,x  \right>=0$ for all $x\in W^\perp$.
If $w\not\in W$, then $w=v+v'$ where $v\in W$ and $v'\in W^\perp\setminus \{0\}$. Then $$0=\left<w,x  \right>=\left<v,x \right>+\left<v',x \right>=\left<v',x  \right>$$
for all $x\in W^\perp.$ However, $v'\in W^\perp$  and  $\left<v',v'  \right>\neq 0$ so it is a contradiction.
A: The converse is trickier, and not necessarily true if $W$ is not finite-dimensional (actually, it is true if and only if $W$ is topologically closed in $V$). So, any proof will have to make use of the finite-dimensionality assumption.
Suppose $(e_1, \ldots, e_n)$ is an orthonormal basis for $W$. Suppose $x \in (W^\perp)^\perp$. Consider the projection $w$ of $x$ onto $W$, specifically:
$$w = \langle x, e_1\rangle e_1 + \ldots + \langle x, e_n \rangle e_n \in W.$$
Then, $x - w$ is perpendicular to each $e_i$, since
\begin{align*}
\langle x - w, e_i\rangle &= \langle x, e_i \rangle - \sum_{j=1}^n\Big\langle \langle x, e_j \rangle e_j, e_i \Big\rangle \\
&= \langle x, e_i \rangle - \sum_{j=1}^n \langle x, e_j \rangle \langle e_j, e_i \rangle \\
&= \langle x, e_i \rangle - \sum_{j=1}^n \langle x, e_j \rangle \delta_{ij} \\
&= \langle x, e_i \rangle - \langle x, e_i \rangle = 0.
\end{align*}
Since the linear map $\langle \cdot, x - w\rangle$ is constantly $0$ on the basis of $W$, it is constantly zero on all of $W$, hence $x - w \in W^\perp$.
But then, since $x \in (W^\perp)^\perp$, we have $\langle x - w, x \rangle = 0$, and since $\langle x - w, w \rangle = 0$,
$$\|x - w\|^2 = \langle x, x - w \rangle - \langle w, x - w\rangle = 0,$$
i.e. $x = w \in W$, completing the proof.
