The Stacks project

Proof. Let $r \in R$ with $s(r) = u$ and $t(r) = u'$. If $u \leadsto u'$ then we can find a nontrivial specialization $r \leadsto r'$ with $s(r') = u'$, see Schemes, Lemma 26.19.8. Set $u'' = t(r')$. Note that $u'' \not= u'$ as there are no specializations in the fibres of a finite morphism. Hence we can continue and find a nontrivial specialization $r' \leadsto r''$ with $s(r'') = u''$, etc. This shows that the orbit of $u$ contains an infinite sequence $u \leadsto u' \leadsto u'' \leadsto \ldots$ of specializations which is nonsense as the orbit $t(s^{-1}(\{ u\} ))$ is finite. $\square$