no, this is wrong. it is like you are using the converse of an implication that is not true. for example, this proof could show that the real numbers are countable! since the naturals are a subset of the reals that are countable. be careful of how you read theorems, they are not always true going backwards. the theorem says if you are given a countable set, then all its subsets will be countable, not if given any set that is countable, then any set containing it will be as well.

right.(2) The other way I thought of proving this was:

Let A and B be countable sets. This means A ~ N and B ~ N (N the set of naturals). A ~ N implies there is a bijection from A to N, and B ~ N implies there is a bijection from B to N. So now I have to show there is a bijection from A x B to N, right?

it suffices to show that is countable (why?)

Now, construct the function given by

where

now show that this function is a bijection (one-to-one and onto)

an alternate way would be to draw a grid (which is usually a nice way to visualize a cross-product of countable sets). where the elements of one set are the first column and the elements of the other form the first row, then the cells in between are the ordered pairs formed from the elements of their respective rows and columns. find a way to traverse this grid to come up with a function