The collection of possible large sets is analytic, namely: $\{A\subset \omega: \forall i<\omega\ \exists U_i\subset T_i \text{ $U_i$ is perfect and } f\restriction \bigcup_{n\in A}\Pi_{i<\omega} T_i(n) \text{ is constant}\}$ (here we can assume the length of the roots of $T_i$ goes to infinity so the coloring $f$ is coded by a real). The finish by a theorem of Mathias: if $P\subset [\omega]^\omega$ is analytic, and $U$ is Ramsey, then there exists $A\in U$ such that $[A]^\omega\subset P$ or $[A]^\omega\cap P =\emptyset$. The first option must appear by density of the set under $\subseteq$.

I saw this argument first from Olga Yiparaki's thesis: On some tree partitions.

To trigger another problem, it may be interesting to think about what happens if the ultrafilter is merely a P-point. Then I think we need to get to the nitty-gritty of Laver's proof.

The collection of possible large sets is analytic, namely: $\{A\subset \omega: \forall i<\omega\ \exists U_i\subset T_i \text{ $U_i$ is perfect and } f\restriction \bigcup_{n\in A}\Pi_{i<\omega} U_i(n) \text{ is constant}\}$ (here we can assume the length of the roots of $T_i$ goes to infinity so the coloring $f$ is coded by a real). The finish by a theorem of Mathias: if $P\subset [\omega]^\omega$ is analytic, and $U$ is Ramsey, then there exists $A\in U$ such that $[A]^\omega\subset P$ or $[A]^\omega\cap P =\emptyset$. The first option must appear by density of the set under $\subseteq$.

I saw this argument first from Olga Yiparaki's thesis: On some tree partitions.

To trigger another problem, it may be interesting to think about what happens if the ultrafilter is merely a P-point. Then I think we need to get to the nitty-gritty of Laver's proof.