It's a known fact that $\mathrm{Ann}(S^\circ)=S$, where $S$ is a subspace of a finite dimensional vector space $V$. I'll include the definitions for the sake of completeness, since $\mathrm{Ann}(S)$ (sometimes notated $^\circ S$ and called the pre-annihilator or joint kernel of $S$) hasn't got a more or less standard definition (for more on this, check this MO thread).
Given a vector space $V$ and a subspace $S\subseteq V$, we define:
$S^\circ=\{\varphi\in V^\star:\varphi(S)=0\}$
Similarly, given a subspace $S^\star\subseteq V^\star$, we define:
$\mathrm{Ann}(S^*)=\{v\in V:\varphi(v)=0\,\,\forall\varphi\in S^\star\}$
On the other hand, if we're dealing with an infinite dimensional vector space, we can only affirm that $S\subseteq \mathrm{Ann}(S^\circ)$. For instance, the inclusion is strict if we work under a normed vector space and choose $S\neq V$ to be any dense subspace (it follows from continuity that any linear functional that vanishes at $S$ must be identically $0$, and therefore $S\subseteq\mathrm{Ann}(S^\circ)=\mathrm{Ann}(\{0\})=V$).
However, a friend of mine sketched a proof that demonstrates that the equality always holds (regardless of the dimension of $V$), and I can't seem to find a hole in neither his argument or the one showed above. It goes like this:
Given $S$, pick any basis $B$ of $S$ and extend it to a basis of $V$ by adding linearly independent vectors outside of $S$, thus forming a basis $B'$ of V. Then, we can define a linear functional $\varphi$ such that $\varphi(v)=0$ for every $v\in B$ and $\varphi(w)=1$ for every $w\in B'\setminus B$. Then, since $\varphi\in S^\circ$, there's no vector $v\in V\setminus S$ such that $v\in\mathrm{Ann}(S^\circ)$. Therefore $S\supseteq\mathrm{Ann}(S^\circ)$, and the other inclusion is easily proved; thus, the equality holds.