**Proof.**
Proposition 101.49.3 tells us that $|\mathcal{X}|$ is sober. Of course it is also irreducible and hence has a unique generic point $x$ (by the definition of sobriety). Proposition 101.29.1 shows the existence of a dense open $\mathcal{U} \subset \mathcal{X}$ which is a gerbe over an algebraic space $U$. Then $U$ is a decent algebraic space by Lemma 101.48.5 (and the fact that $\mathcal{U}$ is decent by Lemma 101.48.3). Since $|U| = |\mathcal{U}|$ we see that $|U|$ is irreducible. Finally, since $\mathcal{U}$ is reduced the morphism $\mathcal{U} \to U$ factors through $U_{red}$, see Properties of Stacks, Lemma 100.10.3. Now since $\mathcal{U} \to U$ is flat, locally of finite presentation, and surjective (Lemma 101.28.8), this implies that $U = U_{red}$, i.e., $U$ is reduced (small detail omitted). It follows that $U$ is an integral algebraic space, see Spaces over Fields, Definition 72.4.1. Then finally, we may replace $U$ (and correspondingly $\mathcal{U}$) by an open subspace and assume that $U$ is an integral scheme, see discussion in Spaces over Fields, Section 72.4.
$\square$

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