@AntonGeraschenko Even in the everybody right/wrong variant, there is a strategy for any cardinality of colors.

Does that guarantee the intervals each have at least $\kappa$ points? In any case, this issue is more subtle than I thought. I'm not sure my claim is true.

I'll write the full argument in the answer.

For any interval with $\kappa^+$ elements, there is a cut such that one side has $\kappa^+$ elements and the other has at least $\kappa$ elements. Recursively cut out an interval with $\kappa$ elements while leaving $\kappa^+$ elements. At the $\alpha$th stage, for limit $\alpha<\kappa,$ the $(\alpha_{\xi}, \beta_{\xi})$ constructed thus far partition the remaining $\kappa^+$ elements into less than $\kappa$ many intervals, one of which has $\kappa^+$ elements, and use that interval to continue the construction.

@MonroeEskew I think I'm using variables inconsistently with the original question. In my casework, $\kappa$ is representing the smaller cardinal (what $\lambda$ is in the problem statement). I'll clarify in my answer.

Yeah, the motivation is shaky. But something nice about the new formulation is that it's a weakening of one characterization of weakly compact cardinals (uncountable $\kappa$ is weakly compact iff $\kappa^*$ or $\kappa$ embeds into every ordering on $\kappa$).

Trivial counterexample: $\omega^*$ ($\omega$ reversed) does not embed into $\omega_1.$

The infinitely many chains might be extraneous. They're an artifact from a previous model. There should be models of ZC+TC without definable infinite sets. I'll email you a sketch I have in mind.

I'm confused by the Claim. It seems to me that there are exactly $2^{\aleph_0}$ equivalence classes.