If $V$ is a finite dimensional vector space then for any two subspaces $U,W$ show that $(U\cap W)^0=U^0+W^0.$ If $V$ is a finite-dimensional vector space then for any two subspaces $U, W$ show that $(U\cap W)^0=U^0+W^0.$ Here $W^0$ means the annihilator of $W.$
My Attempt: We know that $\dim (U\cap W)+\dim (U\cap W)^0=\dim V$ and that $\dim (U+ W)=\dim U+\dim W-\dim (U\cap W).$ So we get that $$\dim U+\dim W-\dim (U\cup W)+\dim (U\cap W)^0=\dim V$$
$$-\dim U^0-\dim W^0+\dim (U\cup W)^0+\dim (U\cap W)^0=0.$$
We also know that $\dim(U^0+W^0)=\dim U^0+\dim W^0-\dim (U\cap W)^0$ and so we get that $$\dim (U^0+W^0)=\dim (U\cap W)^0.$$ Now we have to show that one is contained in the other and then we are done. If we pick $\sigma \in U^0+W^0$ we observe that $\sigma=\phi_u+\psi_w.$ If any $\alpha\in U\cap W$ then $\sigma(\alpha)=\phi_u(\alpha)+\psi_w(\alpha)=0+0=0$ and so $\sigma\in (U\cap W)^0.$ Hence they are equal. Is this proof correct? 
 A: Can we first prove that $(U+W)^0=U^0\cap W^0$ (i.e. a similar identity where $+$ and $\cap$ have switched places)?

If $f\in (U+W)^0$, then $f(v)=0$ for all $v\in U+W\supset U, W$, so $f\in U^0$ and $f\in W^0$, i.e. $f\in U^0\cap W^0$.
Conversely, if $f\in U^0\cap W^0$, then $f\in U^0$ and $f\in W^0$, so for $v\in U+W$, $v=u+w$ for some $u\in U, w\in W$, we have $f(v)=f(u)+f(w)=0+0=0$, thus $f\in (U+W)^0$

Note this is similar to the 2nd part of your proof. It is just that, when you switch $+$ and $\cap$ you can prove a stronger statement (equality rather just one-sided inclusion). You don't need any dimensionality here.
For the rest, we do similar as in your first part of the proof, using dimensional analysis:

$$\begin{align}\dim(U\cap W)^0 & = \dim V-\dim(U\cap W) \\ & = \dim V-(\dim U + \dim W - \dim(U+W)) \\ & = \dim V-(\dim V-\dim{U^0} + \dim V-\dim{W^0}-\dim V + \dim(U+W)^0) \\ & = \dim{U^0}+\dim{W^0}-\dim(U^0 \cap W^0) \\ & = \dim(U^0+W^0)\end{align}$$

Now, by using your (correctly proven) inclusion $U^0+W^0\subseteq (U\cap W)^0$, we reach the desired conclusion.
A: You should fix $U\cup W$ into $U+W$. With this correction, your argument is good. 
Here's a shorter one.
If $V$ is finite dimensional, we have $U^{00}=U$ for all subspaces $U$. Hence we can prove that $U=W$ if and only if $U^0=W^0$.
Thus your statement is equivalent to $(U\cap W)^{00}=(U^0+W^0)^0$, that is, setting $U_1=U^0$ and $W_1=W^0$,
$$
U_1^0\cap W_1^0=(U_1+W_1)^0
$$
This relation holds for all subspaces even in infinite dimensional spaces and is easy to prove.
