Does the "special continuum hypothesis" imply the axiom of choice? In $\mathsf{ZF}$ set theory, does the "special continuum hypothesis" imply the axiom of choice, or is the axiom of choice independent of it?
Here, by the "special continuum hypothesis" we mean the statement that every infinite set of real numbers either has the same cardinality as the set of reals, or the same cardinality as the set of naturals.

There are no implications whatsoever.
As a general rule, if $\varphi$ is a statement which only deals with a single definable set (e.g. the natural numbers, or the real numbers), then it will not imply the axiom of choice.
More correctly, if $\varphi$ is a statement such that for some $\alpha$, $\varphi$ holds if and only if $(V_\alpha,\in)\models\varphi$, then $\varphi$ cannot imply the axiom of choice. Moreover, if $\sf ZFC$ does not prove $\varphi$, then neither will $\sf ZF$ (since we only removed axioms).
The reason is that given a model of $\sf ZFC$, we can always violate choice only above a given rank. So we can violate $\sf AC$ so high, that a statement like "there is an intermediate cardinal between $\aleph_0$ and $2^{\aleph_0}$" (something decided on $V_{\omega+5}$ or so, depending on how you encode functions) cannot possibly imply the axiom of choice, but it will not follow from it either.
More concretely, the continuum hypothesis has many different formulations which are non-equivalent in $\sf ZF$. And stating that there is no intermediate cardinal will not even imply that $2^{\aleph_0}=\aleph_1$.
Relevant Links.


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*Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF

*The Continuum Hypothesis & The Axiom of Choice

*How to formulate continuum hypothesis without the axiom of choice?

*What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that...

*Failure of Choice only for sets above a certain rank
