Question: Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$.

Prove that $a = b$, then prove that $a$ is either $1$ or $2$.

I was thinking using the theorem:

> 
> If $n|a$ and $n|b$,then $n|(ax+by)$ for any $x,y\in \mathbb Z$.
> 
> 
> 

Then I can derive that $ab$ divides $a$ and $b$. I could not go further. However, this doesn't seem like can help me solve the question I had.

Answer: 
Your supposition is mistaken; $\wp(X\times X)$ is not the same as $\wp(X)\times\wp(X)$.

If you want actual equality as sets, then, as Rahul as mentioned, the elements of $\wp(X\times X)$ are sets of pairs of elements of $X$, while the elements of $\wp(X)\times\wp(X)$ are pairs of sets of elements of $X$.

If you just want them to be naturally in bijection with one another... well, that doesn't work either. In general, $\wp(X)\times\wp(Y)$ is naturally in bijection with $\wp(X\amalg Y)$ (here $\amalg$ is disjoint union), which is rather different.

Other people have already pointed out how to show that $X\times X$ is discrete, so I'll skip that...
Careful: if $(X, \tau\_X)$ and $(Y, \tau\_Y)$ are topological spaces, then the product topology on $X \times Y$ is not $\tau\_X \times \tau\_Y$. Rather, it is the topology *generated by* $\tau\_X \times \tau\_Y$, by which I refer to the set $\{ U \times V \mid U \in \tau\_X,\ V \in \tau\_Y \}$, rather than the cartesian product of $\tau\_X$ and $\tau\_Y$.

That is, an open set in $X \times Y$ is one which is a finite intersection of, or arbitrary union union of, sets of the form $U \times V$ for $U \in \tau\_X$ and $V \in \tau\_Y$. In fact, since unions commute with cartesian products, the product topology is actually just the set of unions of sets of the form $U \times V$ as above.

Thus
$$\tau\_{X \times Y} = \left\{ \bigcup\_{i,j} U\_i \times V\_j \ \middle|\ U\_i \in \tau\_X \text{ and } V\_j \in \tau\_Y \text{ for all } i,j \right\}$$

The discrete topology on $X \times Y$ is indeed given by $\mathcal{P}(X \times Y)$. As such, what you need to prove is not that $\mathcal{P}(X) \times \mathcal{P}(Y) = \mathcal{P}(X \times Y)$, which is false in general, but rather that every element of $\mathcal{P}(X \times Y)$ can be expressed as a union of sets of the form $U \times V$ for $U \subseteq X$ and $V \subseteq Y$.

*Edit: For your question, read all '$Y$'s in the last paragraph as '$X$'s.*