If $S$ is a subset of a Hilbert space $H$ then $S^\perp$ is closed? Is this a true statement? (I found it as a theorem in a paper)

If $S$ is a subset of a Hilbert space $H$ then $S^\perp$ is closed.

If it were true then $(S^\perp)^\perp$ would be closed, that is $S$ would be closed. How can two complementary sets be closed?
So to me it seems it is false.
 A: If $s \in S$, then $\mathcal{N}_s  = \{ x \in H : \langle x,s \rangle = 0 \}$ is the null space of a continuous linear functional, which is the inverse image of $\{0\}$ under this continuous linear functional. Hence $\mathcal{N}_s$ is closed, as is the intersection
$$
             S^{\perp} = \bigcap_{s\in S}\mathcal{N}_s.
$$
A: It's true that $S^\perp$ is closed. This is trivial from the definition.
This does not imply that $S$ is closed, because in general $S\ne S^{\perp\perp}$. In fact $S=S^{\perp\perp}$ if and only if $S$ is a closed subspace of $H$; in general $S^{\perp\perp}$ is the closure of the span of $S$.
A: Yes, this is true. Indeed, let $t_n$ be a sequence in $S^{\perp}$ and let $t$ be its limit. We want to show that $t \in S^{\perp}$. 
By the continuity of the inner product, we have that, for all $s \in S$ 
$$
\langle t, s \rangle = \lim_{n \to \infty} \langle t_n, s \rangle=0
$$
so that $t \in S^{\perp}$.
The equality $(S^{\perp})^{\perp}=S$ only holds if $S$ is a closed subspace.
