Proof that $S^\perp$ is a subspace of a vector space $V$

Question

Just doing some review for a final exam and would like some feed back on the following proof if anyone would like to help me out. First the premise.

Let $V$ be a finite dimensional inner product space and $S$ be an orthogonal set in $V$. Prove that $S^\perp$ is a subspace of $V$ such that $S^\perp = \{ v \in V:\langle v,x\rangle = 0,\; \forall x\in S\}$.

Proof:

1. Since $0\in V$ then $\langle0,x\rangle = 0 $ for all $x\in S$, hence $0\in S^\perp$.

2. Let $u, v \in S^\perp$ and $c \in F$ such that $\langle cu\ ,x\rangle = 0 $ and $\langle v \ , x\rangle = 0$ for all $x \in S$. Observe that, $$\langle cv + u, x\rangle\ =\ \langle cv,x\rangle + \langle u,x\rangle\ = \ c\langle v,x\rangle + \langle u,x\rangle\ =\ 0 + 0\ =\ 0$$ Hence $cv + u \in S^\perp$.

Therefore $S^\perp$ is a subspace of $V$.