So we have to prove (axiom of choice is given) that $\aleph_1$ which is the next aleph after $\aleph_0=\omega$ satisfies $$\aleph_1\leq A$$ given that $A$ is uncountable.
Here was my reasoning. The axiom of choice is equivalent to saying that any two cardinalities are comparable. Thus, assume that $A<\aleph_1$, by definition of the alephs, we must have that $|A|=\omega$ which means that $A$ is equinumerous with the naturals, the definition of countable. Hence, by contrapositive we are done.
Is this argument right? The professor said on the midterm we need to use transfinite recursion, but I dont see how.