The Stacks project

Proof. Let $T = \overline{\{ x\} } \subset |\mathcal{X}|$ be the closure of $x$. By Properties of Stacks, Lemma 100.10.1 there exists a reduced closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $T = |\mathcal{X}'|$. Note that $\mathcal{I}_{\mathcal{X}'} = \mathcal{I}_\mathcal {X} \times _\mathcal {X} \mathcal{X}'$ by Lemma 101.5.6. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is quasi-compact as a base change, see Lemma 101.7.3. Therefore Proposition 101.29.1 implies there exists a dense open substack $\mathcal{U} \subset \mathcal{X}'$ which is a gerbe. Note that $x \in |\mathcal{U}|$ because $\{ x\} \subset T$ is a dense subset too. Hence a residual gerbe $\mathcal{Z}_ x \subset \mathcal{U}$ of $\mathcal{U}$ at $x$ exists by Lemma 101.28.15. It is immediate from the definitions that $\mathcal{Z}_ x \to \mathcal{X}$ is a residual gerbe of $\mathcal{X}$ at $x$. $\square$