Basically, you just need to show that .
You can, if you wish, note that if |X| is odd then the complement of every even set us a unique odd set, and your done (taking the complement of X with X to be the empty set, and even set). This is just an observation, it isn't needed to prove the result per se.
Now, if |X|=n is then induct on n. n=1 can be checked manually (note that the empty set is take to be even).
Let where is some arbitrary element. The result holds for Y. Can you conclude that the result thus holds for X? (The subsets of X are the subsets of Y together with all the subsets of Y with x (as in, if the )).