Is the axiom of choice used here?

Question

I was reading the following How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles?

The answer by "user4594" basically considers a k-cell with rational endpoints, and I am guessing that when he says that "E would be the union of these cells" the axiom of choice is needed because for each x there may be more than one k-cell with rational endpoints.

Am I correct?

Answer 1

You could actually take all rectangles with rational endpoints containing $x$, since there are at most countably many of them, so you don't need choice.

Anyway, choice on subsets of a countable set is provable without the full Axion of Choice.

Answer 2

Without the axiom of choice the rational numbers can be well-ordered, since they are countable. Therefore the set of $k$-rectangles with rational endpoints is countable.

So we don't need the axiom of choice to really choose from this collection.