 Solution

False. In ${\mathbb{R}}^2$ let $U=\mathrm{span}(1,1)$, $W=\mathrm{span}(1,0)$ and $X=\mathrm{span}(0,1)$. Then $W+X={\mathbb{R}}^2$, so $U\cap (W+X)=U$, but $U\cap W=U\cap X=\{0\}$ and their sum is zero. Thus intersection need not distribute over the sum of vector subspaces. The right side is always contained in the left; this example makes the inclusion strict.