The whole question looks like-
Let, $S\subseteq V^*$ and define $\text{Ann}(S)=\{v\in V|f(v)=0\ \forall f\in S\}$. Prove that, $\text{Ann}(W_1)+\text{Ann}(W_2)=\text{Ann}(W_1\cap W_2)$ where $W_1,W_2$ are two subspaces of $V^*$.(Here $V^*=\mathcal{L}(V, F)=$set of all linear maps from $V$ to $F$)
I have just proved one part of the equality $\text{Ann}(W_1)+\text{Ann}(W_2)\subseteq\text{Ann}(W_1\cap W_2)$ which goes as follows-
Let, $v\in \text{Ann}(W_1)+\text{Ann}(W_2)\implies v=w_1+w_2$ where $w_1\in W_1, w_2\in W_2 \implies f(w_1)=0\ \forall f\in W_1, g(w_2)=0\ \forall g\in W_2$.
Now let $f\in W_1\cap W_2\implies f(v)=f(w_1+w_2)=f(w_1)+f(w_2)=0+0=0$
Thus we get $f(v)=0\ \forall f\in W_1\cap W_2\implies v\in\text{Ann}(W_1\cap W_2)$. Hence, $\text{Ann}(W_1)+\text{Ann}(W_2)\subseteq\text{Ann}(W_1\cap W_2)$.
But I can't establish the other part i.e. $\text{Ann}(W_1\cap W_2)\subseteq\text{Ann}(W_1)+\text{Ann}(W_2)$.
Any anybody give me a proper way out? Thanks for assistance in advance.