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.
 A: 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 $u =u_1 \in S.$
A: First off, the question asks for an "iff", so you should expect your proof to be split into two parts, one for each of the two implications.
One of those two implications is much easier than the other (can you see which one?).
So your proof should look roughly like this:

Let $S$ be a subset of $V$. Assume $(S^\perp)^\perp = S$. Then [...]. Thus $S$ is a vector subspace of $V$.
Let $S$ be a vector subspace of $V$. Then [...]. Thus $(S^\perp)^\perp = S$.

But note that in the second part, you are proving a set equality. So you should expect the second part of your proof to be split in two parts: assuming $S$ is a vector subspace, show $S \subseteq (S^\perp)^\perp$ and show $(S^\perp)^\perp \subseteq S$.
