I was thinking if we define a set such that the $n$th digit is a random natural number less than or equal to $n$ in such a way that we favor the numbers 1-3 highly, and then has an even distribution up to $n$, and we remove duplicates only after determining them. The elements are the numbers themselves, not the probability distribution.
We end up with a clearly infinite set because there's no definite largest number in the set, but it's also not countable since we defined the probability to highly favor small numbers, so without an infinite amount of processing, there are only small numbers, but we can't say that there are no large numbers.
This hasn't broken any axioms in ZF as far as I'm aware. If we go through them, it has extensionality, it's regular, it works with subsets, it's pairable, unions work with it. I honestly don't understand the axiom of replacement well enough but I'm almost certain it works out based on what I've read, but I may be wrong. This exists in a world with infinite sets. The axiom of power set is interesting here, since it's not well defined but it still has subsets, I think we can construct a set of all its subsets and that may or may not be countable. It's definitely well ordered since we started with a countably infinite set and this is a strict subset of that set, so we can just use the same ordering.
I know I messed up somewhere, and the axiom of choice breaks this since you can't pick any element other than the first one since it may be a duplicate, so you don't get to choose which element to pick. I'm fascinated to find out where I messed up here. Proof verification is a tag I listed, but this is less of a proof, more so evidence.
This set is essentially $\{1,2,3,a_1,a_2,\cdots\}$ but we can't say that $a_1\ne a_2$ or $a_1\leq a_2$ or even that $a_n$ actually exists, so we can't say it's countable since we don't know the elements. However, there's no largest element, so it's infinite.
