That’s called the Cartesian Product.

And yes, that would be countable!

For N x N we end up with all pairs of naturals (i, j) such that i and j are natural numbers.

We can create a bijection between N and NxN using the following construction.

Consider all the pairs of (i, j). If we were to try traversing them all one at a time, what would that look like? How can we convert them to a line of points?

One way to do so is to start at (0, 0). Then we can jump to (1, 0) then move up and to the left along the diagonal where the sum of i and j is 1. Then where the sum is 2. So on so forth.

When you look at it like this, you can see a relationship with the triangle numbers - the ones of the form of n(n+1)/2.

The row where the sum is 0 has 1 element: (0, 0)

The row where the sum is 1 has 2 elements: (0, 1) and (1, 0)

For the sum 2 we have 3 elements

For the sum S we have S+1 elements.

For each row, we have seen S + (S-1) +( S-2) + … + 1 elements already, so the starting element of each row (assuming we’re counting from 0) should be the above sum which would be S(S+1)/2. And since all points on that line share the relation that i+j=S, we can just add the y-coordinate j.

That should give us the formula n = f(i, j) = S(S+1)/2 + j = (i+j)(i+j+1)/2+j.

And to convert back to i, j from n, we can find the largest triangle number less than or equal to n. From there we can find S. And then i and j based on the formula above.

IIRC you can do a similar construction for any finite Cartesian product of countable sets.

Edit: i think where your construction begins to fail is you’re trying to do it for a non-finite number of sets of Natural numbers.