If $u \colon R \to U$ is a monomorphism, then since $P$ is injective, there is a morphism $h \colon U \to P$ such that $i=hu$.

But by an earlier theorem $i \colon R \to P$ is an irreducible morphism, meaning that in the factorization $i=hu$ either $u$ is a section or $h$ is a retraction. We have assumed $u$ is not a section, so $h$ is a retraction and $u$ factors through $P$.