The standard construction of a Vitali set only involves making choices from countable subsets of $\mathbb{R}$ (specifically, from sets of the form $(r+\mathbb{Q}) \cap [0,1]$). It is well known that ZF (assuming consistent) does not prove the existence of a non-measurable subset of $\mathbb{R}$, hence it doesn't prove the existence of a Vitali set, and thus doesn't prove your restricted choice principle.

The standard construction of a Vitali set only involves making choices from countable subsets of $\mathbb{R}$ (specifically, from sets of the form $(r+\mathbb{Q}) \cap [0,1]$). It is well known that ZF (assuming consistent) does not prove the existence of a non-measurable subset of $\mathbb{R}$, hence it doesn't prove the existence of a Vitali set, and thus doesn't prove your restricted choice principle.

Consequences of the Axiom of Choice is useful for this sort of thing. According to this, form 85 (every collection of countable sets has a choice function) is true while form 79 (every collection of sets of reals has a choice function) is false in model $\mathcal{M} 1$, which I believe is Cohen's original model of ~AC. Thus your countable choice principle for the reals does not imply full choice for the reals.

The standard construction of a Vitali set only involves making choices from countable subsets of $\mathbb{R}$ (specifically, from sets of the form $(r+\mathbb{Q}) \cap [0,1]$).

The standard construction of a Vitali set only involves making choices from countable subsets of $\mathbb{R}$ (specifically, from sets of the form $(r+\mathbb{Q}) \cap [0,1]$). It is well known that ZF (assuming consistent) does not prove the existence of a non-measurable subset of $\mathbb{R}$, hence it doesn't prove the existence of a Vitali set, and thus doesn't prove your restricted choice principle.