Hint for a more "algebraic" solution:
$$ A \cap (B \setminus A) = A \cap (B \cap A^c) = A \cap (A^c \cap B) = (A \cap A^c) \cap B = \ldots.$$
By the way, the problem seems a bit strange for asking you to show that a given set is a subset of the empty set. In general, to show $X \subseteq Y$ you would let $x \in X$ and prove that $x \in Y$. So to show that $X \subseteq \emptyset$ you would let $x \in X$ and then prove that $x \in \emptyset$. Of course "$x \in \emptyset$" is absurd, so this method amounts to letting $x \in X$ and proving absurdity (a contradiction.) Note that a contradiction implies everything, including the desired conclusion that $x \in \emptyset$.
An equivalent but more natural formulation of the question would simply ask you to show that the given set was empty. After all, the only subset of the empty set is the empty set itself.