Well since you wanted to use the contrapositive for A intersects B = empty -> B is a subset of X\A I'll explain that way. Let [itex]B=\{b_1,b_2,...\}[/itex]. If B is empty what we want is vacuously true since the empty set is a subset of every set (and thus B can never not be a subset of [itex]X \setminus A[/itex]). If [itex]B\not\subset X\setminus A[/itex] then at least one [itex]b_i \in B[/itex] is [itex]\in X\setminus (X \setminus A) = A[/itex] which implies A and B have these elements in common.
You can prove the other way in a similar fashion. My personal suggestion is contradiction for the other way.
On this note, you have another question that's pretty similar. Your questions really boil down to choosing elements of certain sets and then showing by logic that they must/must not exist in other sets. Try to proceed like this in your other question also.