 Solution

False. Take $A=\{0,1\}$, $B=\{0\}$ and the constant function $f(a)=0$. For the subsets $X=\{0\}$ and $Y=\{1\}$, the intersection $X\cap Y$ is empty, so $f(X\cap Y)=\varnothing$, whereas the intersection of their images of a function is $f(X)\cap f(Y)=\{0\}$. In general the inclusion $f(X\cap Y)\subseteq f(X)\cap f(Y)$ holds, but a common image value can arise from different points of $X$ and $Y$.