4 Answers
4

The first implication holds because $A$ and $B^\prime$ cannot have any elements in common (this is what $A \cap B^\prime = \emptyset$ means). The second implication holds because any element you pick must either belong to a given set or belong to its complement, but not both.
–
Austin MohrMar 22 '12 at 6:18

Since you are asking for the steps for proving the statement, I would like to say that write down the definitions of
$$
A\cap B^c=\emptyset
,$$ $$B^c,$$
and
$$
A\subset B
$$
first. And then try to go on. (Here $B^c$ is your $B'$.)

Suppose not, then $x \in B'$. We have $x \in A \cap B' = \emptyset$ which is not possible.

Hence $A \subseteq B$.

Edit: I answered this question to test the proof approach and technique I learnt from school years ago. Would anybody point out anything that is wrong or inappropriate in the proof so that I can improve them? Thanks.