How do I prove propositions involving power sets and cartesian products? In an earlier question I asked regarding how to solve specific propositions involving set unions/intersections etc. What helped greatly is the use of axioms and rules that I could use to prove the propositions. However, when it comes specifically to propositions involving power sets and cartesian products, I'm having a much harder problem because I can't seem to use any axioms. 
For example I need to prove whether this statement is true (where A and B are sets and P is power set):
if      P(A)    ⊆   P(B)        then        A   ⊆   B
Or for example I have things involving both power sets and cartesian products:
P(A X   B)  ⊆       P(A)    X   P(B)
Again, what I'm lacking is the actual process, possibly step-by-step, to do to prove these kinds of statements. For the first one, what I would normally do is find a way to break down the "left hand side" and see whether I can isolate A or B so that I can replace them on the right hand side. But nothing comes up. Same goes for the second statement. How do I break these into tinier things that I can use to prove the other part?
 A: If $$A\times B=\{(a,b):a\in A, b\in B\},$$
then here are hints for both:


*

*$P(A)\subset P(B)$ then for any $a\in A$, $\{a\}\in P(B).$

*$ \{(a,b)\}\in P(A\times B)$ . Can you find him in $P(A)\times P(B)$?
A: First statement: since you want to prove that $A\subseteq B$, start taking any $x\in A$. You must show that $x\in B$. You can (and, in this case, must) use the hypothesis, of course. So think in a subset of $A$ using what you have, namely, the element $x$. Since $\mathcal P(A)\subseteq\mathcal P(B)$, this subset is also a subset of $B$. This should imply that $x\in B$.
Second statement: It is false, since $\emptyset\in\mathcal P(A\times B)$ but $\emptyset$ is not a pair.
A: For the statement $P(A) \subseteq P(B)$, consider the fact that:
$$A \in P(A)$$
In general, when proving things about power sets, it's often best to go back to the original definition:
$$x \in P(y) \iff x \subseteq y$$
Thus, in the above case we could say $A \subseteq B \iff A \in P(B)$, which gives an obvious direction as to where to use the hypothesis.

When dealing with (binary) Cartesian products, it's often useful to think of rectangles in 2D, each pair consisting of two "coordinates", as it were. Once you get the hang of it, you can do the same with bigger products, eventually arbitrary (infinite) products.
If you draw the situation for $P(A\times B) \subseteq P(A)\times P(B)$ on a piece of paper, you see that it amounts to the assertion that every subset of a rectangle is itself a smaller rectangle. This intuition should help you conjure a counterexample.
