The Stacks project

Proof. Assume the pair $(X, x)$ satisfies the equivalent conditions for Lemma 68.4.2. Let $U$ be a scheme, $U \to X$ étale, and let $u, u' \in |U|$ map to $x$ of $|X|$, and $u' \leadsto u$. We may and do replace $U$ by an affine neighbourhood of $u$. Let $t, s : R = U \times _ X U \to U$ be the étale projection maps.

Pick a point $r \in R$ with $t(r) = u$ and $s(r) = u'$. This is possible by Properties of Spaces, Lemma 66.4.5. Because generalizations lift along the étale morphism $t$ (Remark 68.4.1) we can find a specialization $r' \leadsto r$ with $t(r') = u'$. Set $u'' = s(r')$. Then $u'' \leadsto u'$. Thus we may repeat and find $r'' \leadsto r'$ with $t(r'') = u''$. Set $u''' = s(r'')$, and so on. Here is a picture:

In Remark 68.4.1 we have seen that there are no specializations among points in the fibres of the étale morphism $s$. Hence if $u^{(n + 1)} = u^{(n)}$ for some $n$, then also $r^{(n)} = r^{(n - 1)}$ and hence also (by taking $t$) $u^{(n)} = u^{(n - 1)}$. This then forces the whole tower to collapse, in particular $u = u'$. Thus we see that if $u \not= u'$, then all the specializations are strict and $\{ u, u', u'', \ldots \} $ is an infinite set of points in $U$ which map to the point $x$ in $|X|$. As we chose $U$ affine this contradicts the second part of Lemma 68.4.2, as desired. $\square$