So, with the annihilator of a set $U$ defined as Anni($U$) = $\{f \in V^* \: | \: \text{for all $u \in U$: }f(u) = 0\}$, and the canonical isomorphism $\varphi: V \to V^{**}$ given by $\varphi(v)(w^*) = w^*(v)$, I want to know if for a subspace $U$ of $V$ we have $\varphi[U] = $ Anni(Anni($U$)).
So far, I have the inclusion $\varphi[U] \subseteq $ Anni(Anni($U$)), and I'll show how. Let $f \in \varphi[U]$, then for some $u \in U$ we have $f(w^*) = w^*(u)$ for all $w^* \in V^*$. Then for all $w^* \in $ Anni($U$) we have $f(w^*) = w^*(u) = 0$, since $u \in U$ and $w^* \in $ Anni($U$). So then we have $f \in $ Anni(Anni($U$)).
However, I'm struggling with proving Anni(Anni($U$)) $\subseteq \varphi[U]$. The problem I'm having is, for any $f \in $ Anni(Anni($U$)), we only know how $f$ acts on the elements of $U$. For any elements in $V \setminus U$ we don't really know where $f$ maps those elements, so I am not sure if $f \in \varphi[U]$, nor do I know how to prove it is or isn't.
If $V$ is finite dimensional, then we have $\dim(\mathrm{Anni}(U)) = \dim(V) - \dim(U)$.
Thus, we may conclude that $\dim[\mathrm{Anni}(\mathrm{Anni}(U))] = \dim(U)$
From this and your inclusion $\mathrm{Anni}(\mathrm{Anni}(U)) \subseteq \varphi[U]$, it follows that $\mathrm{Anni}(\mathrm{Anni}(U)) = \varphi[U]$.
This statement will generally fail for infinite dimensional spaces, for either generalization of a dual space.
If you take $V^*$ to refer to the algebraic dual of $V$, then $\varphi:V \to V^{**}$ will fail to be an isomorphism whenever $V$ is infinite dimensional.
If you take $V$ to be a topological vector space and take $V^*$ to refer to the continuous dual, then we can construct a counterexample as follows: when $V$ is a Hilbert space, we have $\mathrm{Anni}(\mathrm{Anni}(U)) = \overline{U}$, and we will have $U \subsetneq \overline{U}$ if $U$ is a non-closed subspace of $V$. Every infinite dimensional normed space has a non-closed subspace.