$(U\cap W)^0 = U^0 + W^0$

Let $V$ be a finite dimensional vector space and $U^0$ and $W^0$ denote the annihilators of subspaces $U$ and $W$, respectively. Then I need to show that $(U\cap W)^0 = U^0 + W^0$.

I can show that $U^0 + W^0 \subseteq (U\cap W)^0$ like so: Let $f\in U^0$ and $g\in W^0$. Then, because $U^0, W^0 \subseteq (U\cap W)^0$, we see that if $f\in U^0$ and $g\in W^0$, then for any $v\in U\cap W$ and $a,b \in \Bbb F$, $(af+bg)(v) = af(v) + bg(v) = 0$. Thus $af+bg\in (U\cap W)^0$. But because $f$ and $g$ were arbitrary, this implies that any linear combination of elements of $U^0$ and $W^0$ is an element of $(U\cap W)^0$. I.e. $U^0+W^0 \subseteq (U\cap W)^0$.

But I haven't had any luck proving the other direction. Obviously I should start with $f\in (U\cap W)^0$. But then it looks like I need to somehow break that function into two to show its in $U^0 + W^0$. That's throwing me off.

Since you have already proved that $U^0 + W^0 \subseteq (U\cap W)^0$, it is enough to prove that their dimensions are equal. Now we have $$\dim(U\cap W)^0=n-\dim (U\cap W),$$ and \begin{align}\dim(U^0+W^0) & =\dim (U^0)+\dim (W^0)-\dim (U^0\cap W^0)\\ & =2n-\dim(U)-\dim(W)-\dim (U+W)^0 \\ & = 2n-\dim(U)-\dim(W)-n+\dim(U+W)\\ &= n-\dim (U\cap W).\end{align}

Note that to obtain the second line, we used the equality $U^0\cap W^0=(U+W)^0$, which simply says that a linear functional $f$ vanishes on $U+W$ if and only if it vanishes on $U$ and $W$.