A set S is a subspace of an inner product space $V$ iff $(S^{\perp})^{\perp}=S$? 
$S$ is any set of a finite-dimensional inner product space $V$. Prove or disprove that $S$ is a subspace of $V$ iff $(S^{\perp})^{\perp}=S$.

My approach was to consider a minimum set $B$ of linearly independent vectors which spans $S$. If such $B$ exists then there exists some $v \in B$ and $v \notin S$, but $v \in (S^{\perp})^{\perp}=(B^{\perp})^{\perp}=B$ so contradiction.
What I can't deal with is how can I consider such $B$. Can I just say "Try every set of $V$ and find the minimum?"
Any help is appreciated.

First of all orthogonal complement is a subspace, so if $S=(S^{\perp})^{\perp},$ then $S$ must be a subspace.
Conversely, it is easy to see that $S \subseteq (S^{\perp})^{\perp}.$ For the other containment,
Hint: First show $$V= S \oplus S^{\perp},$$ so for $u \in (S^{\perp})^{\perp},$ there exists $u_1 \in S,u_2 \in S^{\perp}$ such that $u=u_1+u_2.$ It follows $u-u_1 \in S^{\perp}.$ You can also show $u-u_1 \in(S^{\perp})^{\perp}.$ Conclude $u =u_1 \in S.$
