 Halting-stage colouring of triples

Let $K_s$ be an increasing uniformly computable finite approximation to the diagonal halting set. For $a<b<c$, colour the triple $0$ if $K_b\upharpoonright a=K_c\upharpoonright a$, and $1$ otherwise. No infinite homogeneous set for a colouring can have colour $1$, since the finitely many bits below its first element eventually stabilize. Every infinite homogeneous set $H$ of colour $0$ computes the diagonal halting set: choose $n<a<b$ in $H$ and read membership of $n$ in $K_b$. Homogeneity and unboundedness make this answer final. Thus the colouring has no infinite computable homogeneous set.