A set S is a subspace of an inner product space $V$ iff $(S^{perp})^{perp}=S$?

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 $u =u_1 in S.$

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 $(S^perp)^perp subseteq S$.