challenge

Perhaps because a simple counterexample would dispose of it.

Take A = {1}, B={2}.

Then \{1,2\} \in P(AUB) but

\{1,2 \} \notin P(A) \bigcup P(B)

You would see {{1},{2}} but of course you know how to distinguish it from {1,2}

Hmmm....i was able to think of the counter example...but to be frank i was looking for how to "WRITE PROOF" formally!

Well then the proof is staring in your face.

The set A \bigcup{} B is not an element of P(A) \bigcup{} P(B) but does belong to P(A \bigcup{} B)