Answer
$F(A) \cap F(B) \nsubseteq F(A \cap B)
$
$counter\,example$
$X=\left \{ 5,6 \right \},Y=\left \{ 5 \right \}\\
A=\left \{ 5 \right \},B=\left \{ 6 \right \}\\
F:X\rightarrow Y ,\,\,\,F(5)=F(6)=5 \\
F(A) \cap F(B)=F(\left \{ 5 \right \})\cap F(\left \{ 6 \right \})=\left \{ 5 \right \}\\
A \cap B=\left \{ 5 \right \}\cap \left \{ 6 \right \}=\varnothing \\
F(A \cap B)=F(\varnothing )=\varnothing \\
but\,\left \{ 5 \right \}\nsubseteq \varnothing \\
F(A) \cap F(B) \nsubseteq F(A \cap B)
$
Work Step by Step
$X=\left \{ 5,6 \right \},Y=\left \{ 5 \right \}\\
A=\left \{ 5 \right \},B=\left \{ 6 \right \}\\
F:X\rightarrow Y ,\,\,\,F(5)=F(6)=5 \\
F(A) \cap F(B)=F(\left \{ 5 \right \})\cap F(\left \{ 6 \right \})=\left \{ 5 \right \}\\
A \cap B=\left \{ 5 \right \}\cap \left \{ 6 \right \}=\varnothing \\
F(A \cap B)=F(\varnothing )=\varnothing \\
but\,\left \{ 5 \right \}\nsubseteq \varnothing \\
F(A) \cap F(B) \nsubseteq F(A \cap B)
$
