The first inclusion is easy to prove. You have to observe that if $w_1\in W_1\cap W_3$ and $w_2\in W_2\cap W_3$ then for $w=w_1+w_2$ apply that:
$$w_1,w_2\in W_3\Rightarrow w=w_1+w_2\in W_3$$
$$w_1\in W_1, w_2\in W_2 \Rightarrow w=w_1+w_2\in W_1+W_2$$
So $w\in (W_1+W_2)\cap W_3$.
Now for the equality. Those two are equals if $W_1\subseteq W_3$ or $W_1\subseteq W_2$, which can be proven in the same way as before. Suppose without loss of generality that $W_1\subseteq W_3$. Then, take $w\in (W_1+W_2)\cap W_3$.
$$w\in (W_1+W_2)\Rightarrow \exists w_1\in W_1\exists w_2\in W_2 : w=w_1+w_2$$
We have that $w_1\in W_1\subseteq W_3$ and $w\in W_3$, hence $w_2=w-w_1\in W_3$ and so it applies that
$w_2\in W_2\cap W_3$ and $w_1\in W_1=W_1\cap W_3$ which is the desirable.