 Solution

For each hypergraph edge $W$, let $B_W$ be the event that one of its two colours occupies more than $3r/4$ vertices. By colour symmetry and the union bound, the preceding calculation givesAgain the dependency graph joins intersecting hypergraph edges and has $D+1\le r(\Delta +1)$. The assumed degree bound impliesThe symmetric Lovász local lemma gives positive probability that no $B_W$ occurs. Thus there is a two-colouring in which each colour occupies at most $3r/4$ vertices of every hypergraph edge. Equivalently, both colour counts in every hypergraph edge lie between $r/4$ and $3r/4$; no rounding difficulty arises because the counts are integers and the bad events use strict inequalities.