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I have a set $S$ which is countable.

I have defined a subset $U \subset S$ that:

  1. cannot have a cardinality higher than $|S|$ (because it's a subset);
  2. is infinite (because I proved that).

Now I'd like to prove that $U$ is countable and I'd like to do it with the following statement:

If $U$ weren't countable, then I'd have a set for which the cardinality is non-finite and $|U| < \aleph_0$ at the same time, disproving the generalized continuum hypothesis. Given that the continuum hypothesis is independent from ZFC, either $U$ is independent too or it is countable.

Then, with this statement, I'd just have to show that my set is not independent from ZFC (which shouldn't be that hard) and I'm done.

Is it right?

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I think the idea of your proof is way too complicated. First of all, note that a set $X$ is at most countable if there is a surjection $f: \omega \to X$. Since we know that $S$ is countable, there is a surjective map $f: \omega \to S$. Then pick $u \in U$ and define $g: \omega \to U$ such that $g(x) = f(x)$ if $x \in U$ and $g(x) = u$ if $x \notin U$. Since $U \subset S$, we know that $g$ is surjective. Hence $U$ is at most countable. Since you know that $U$ is infinite, we know that $U$ has to be countable, i.e. $|U| = \omega$.

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You can prove, without all the independence, or whatnot, that if $A$ is an infinite set, and $|A|\leq\aleph_0$, then $|A|=\aleph_0$.

Since $U$ is a subset of a countable set, if you prove that $U$ is infinite, you're done.

The continuum hypothesis (generalized or otherwise) has nothing to say about cardinals smaller than $\aleph_0$, since those are all provably finite anyway. And the continuum hypothesis is only relevant for infinite cardinals.

So, your proof is generally on the wrong track.

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