Explain why X is discrete

Thanks to comment and JuanPablo
http://stacks.math.columbia.edu/tag/02G7#comment-2245
http://stacks.math.columbia.edu/tag/02G7#comment-2247

morphisms.tex

@@ -7150,9 +7150,10 @@ \section{Unramified morphisms}
 Conversely, suppose that $X \to \Spec(k)$ is unramified.
 By Algebra, Lemma \ref{algebra-lemma-unramified-at-prime} for every $x \in X$
 the residue field extension $k \subset \kappa(x)$ is
-finite separable. Hence all points of $X$ are closed points
-and even isolated points, see
-Lemma \ref{lemma-algebraic-residue-field-extension-closed-point-fibre}.
+finite separable. Since $X \to \Spec(k)$ is locally
+quasi-finite (Lemma \ref{lemma-unramified-quasi-finite})
+we see that all points of $X$ are isolated closed points, see
+Lemma \ref{lemma-quasi-finite-at-point-characterize}.
 Thus $X$ is a discrete space, in particular the disjoint union
 of the spectra of its local rings. By
 Algebra, Lemma \ref{algebra-lemma-unramified-at-prime} again these