Is it consistent that there are regular cardinals $\kappa < \lambda$, such that $\lambda$ is a successor cardinal and for every coloring $d\colon[\lambda]^2\to\kappa$ there is some $A\subseteq\lambda$ of cardinality $\lambda$ and $\eta < \kappa$ such that $\forall x,y\in A$, $d(x,y)< \eta$?
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If $\lambda\nrightarrow[\lambda]^2_\lambda$ then there is no such $\kappa<\lambda$: modify the coloring that witnesses this by defining it to be $0$ on pairs of ordinals sent to values above $\kappa$. (The original coloring takes on all values on any unbounded subset of $\lambda$, hence the modified coloring takes on all values below $\kappa$ on such a set.)
This rules out $\lambda$ being successor of regular (in fact, it rules out $\lambda$ having a non-reflecting stationary set by a result of Todorcevic).
For successor of singular, the question of whether $\lambda\nrightarrow[\lambda]^2_\lambda$ can fail is open.
Edit #1:
What you ask follows if $\lambda$ is a Jonsson successor of singular, as the failure of your condition at successor of singular gives a Jonsson algebra.
Edit #2:
If you demand that your condition applies to colorings of finite subsets of $\lambda$, then you end up with something equivalent to $\lambda$ being Jonsson.
answered Sep 6, 2015 at 0:59
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$\begingroup$ Going to quit with the edits for the night, but I'm pretty sure you condition (for a given successor of singular $\lambda$) is equivalent to $\lambda\rightarrow[\lambda]^2_\lambda$. Will flesh out the argument tomorrow. $\endgroup$ Sep 6, 2015 at 3:17
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$\begingroup$ Not quiet as easy as I hoped. Still think it's true, but will require some work to get a proof. Will work on it. $\endgroup$ Sep 6, 2015 at 18:43