Answer
$F(A \cap B) \subseteq F(A) \cap F(B) \\
let\,\,y\in F(A \cap B)\Rightarrow y=F(x)\,for\,some\,x\in A\cap B\\
x\in A\cap B\Rightarrow x\in A\,and\,x\in B \\
so\,\,y=F(x)\\
for\,some\,x\in A \Rightarrow y\in F(A)\\
for\,some\,x\in B \Rightarrow y\in F(B)\\
\because y\in F(A)\,and\,y\in F(B)\\
\therefore y\in F(A)\cap F(B)\\
\because y\in F(A\cap B)\Rightarrow y\in F(A)\cap F(B)\\
\therefore F(A \cap B) \subseteq F(A) \cap F(B) \\
$
Work Step by Step
$F(A \cap B) \subseteq F(A) \cap F(B) \\
let\,\,y\in F(A \cap B)\Rightarrow y=F(x)\,for\,some\,x\in A\cap B\\
x\in A\cap B\Rightarrow x\in A\,and\,x\in B \\
so\,\,y=F(x)\\
for\,some\,x\in A \Rightarrow y\in F(A)\\
for\,some\,x\in B \Rightarrow y\in F(B)\\
\because y\in F(A)\,and\,y\in F(B)\\
\therefore y\in F(A)\cap F(B)\\
\because y\in F(A\cap B)\Rightarrow y\in F(A)\cap F(B)\\
\therefore F(A \cap B) \subseteq F(A) \cap F(B) \\
$