Lemma 101.49.1. Let \mathcal{X} be a decent algebraic stack. Then |\mathcal{X}| is Kolmogorov (see Topology, Definition 5.8.6).
Proof. Let x_1, x_2 \in |\mathcal{X}| with x_1 \leadsto x_2 and x_2 \leadsto x_1. We have to show that x_1 = x_2. Let \mathcal{Z} \subset \mathcal{X} be the reduced closed substack with |\mathcal{Z}| equal to \overline{\{ x_1\} } = \overline{\{ x_2\} }. By Lemma 101.48.3 we see that \mathcal{Z} is decent. After replacing \mathcal{X} by \mathcal{Z} we reduce to the case discussed in the next paragraph.
Assume |\mathcal{X}| is irreducible with generic points x_1 and x_2. Pick an affine scheme U and u_1, u_2 \in U and a smooth morphism f : U \to \mathcal{X} such that f(u_ i) = x_ i. Then we find a third point u_3 \in U which is the generic point of an irreducible component of U whose image x_3 \in |\mathcal{X}| is also a generic point of |\mathcal{X}|. Namely, we can simply choose u_3 any generic point of an irreducible component passing through u_1 (or u_2 if you like). In the next paragraph we will show that x_1 = x_3 and x_2 = x_3 which will prove what we want.
By symmetry it suffices to prove that x_1 = x_3. Since x_1 is a generic point of |\mathcal{X}| we have a specialization x_1 \leadsto x_3. By Lemma 101.47.1 we can find a specialization u'_1 \leadsto u_3 in U (!) mapping to x_1 \leadsto x_3. However, u_3 is the generic point of an irreducible component and hence u'_1 = u_3 as desired. \square
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