Non-Ramsey functions $c:[\omega]^\omega\to\{0,1\}$ and the Axiom of Choice

Question

Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$, and let $c:[\omega]^\omega\to\{0,1\}$ be a function. We say that $a\in [\omega]^\omega$ is monochromatic with respect to $c$ if the restriction $c|_{{\cal P}(a)\cap [\omega]^\omega}$ of $c$ to the infinite subsets of $a$ is constant. Moreover, $c$ is said to be Ramsey if there is $a\in [\omega]^\omega$ such that $a$ is monochromatic with respect to $c$.

Using the Axiom of Choice it is possible to construct a non-Ramsey function $c:[\omega]^\omega\to\{0,1\}$.

Question. Does the existence of a non-Ramsey function $c:[\omega]^\omega\to\{0,1\}$ imply some (weak) form of (AC) that cannot be proved in ${\sf (ZF)}$?

Answer 1

The keyword for searching is "infinite exponent partition relation."

The following article, for example, has many interesting results concerning the infinite exponent partition relation $\omega\to(\omega)^\omega$, which asserts that every coloring function has your Ramsey property, and its connection with weak forms of choice.

Kleinberg, E. M.; Seiferas, J. I., Infinite exponent partition relations and well-ordered choice, J. Symb. Log. 38, 299-308 (1973). ZBL0274.04004.

In particular, they prove that under well-ordered choice, which is a weakening of the axiom of choice to well-ordered index sets (e.g. countable choice is an instance of this), then case $\omega\to(\omega)^\omega$ is the only possible case.

Meanwhile, it is mentioned there that Adrian Mathias, in his dissertation, proved that the infinite exponent partition relation $\omega\to(\omega)^\omega$ is consistent with countable choice and indeed with dependent choice. Consequently, if ZF is consistent, those choice principles are insufficient to produce a non-Ramsey function.