The Stacks project

Proof. Choose an affine scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$, see Properties of Stacks, Lemma 100.6.2. Then $|U| \to |\mathcal{X}|$ is continuous, open, and surjective, see Properties of Stacks, Lemma 100.4.7. Hence the quasi-compact opens of $|\mathcal{X}|$ form a basis for the topology. For $W_1, W_2 \subset |\mathcal{X}|$ quasi-compact open, we may choose a quasi-compact opens $V_1, V_2$ of $U$ mapping to $W_1$ and $W_2$. Since $\Delta $ is quasi-compact, we see that

is quasi-compact. Then image of $|V_1 \times _\mathcal {X} V_2|$ in $|\mathcal{X}|$ is $W_1 \cap W_2$ by Properties of Stacks, Lemma 100.4.3. Thus $W_1 \cap W_2$ is quasi-compact. To finish the proof, it suffices to show that $|\mathcal{X}|$ is sober, see Topology, Definition 5.23.1.

Let $T \subset |\mathcal{X}|$ be an irreducible closed subset. We have to show $T$ has a unique generic point. Let $\mathcal{Z} \subset \mathcal{X}$ be the reduced induced closed substack corresponding to $T$, see Properties of Stacks, Definition 100.10.4. Since $\mathcal{Z} \to \mathcal{X}$ is a closed immersion, we see that $\Delta _\mathcal {Z}$ is quasi-compact: first show that $\mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ is quasi-compact as the composition of $\mathcal{Z} \to \mathcal{X}$ with $\Delta $, then write $\mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ as the composition of $\Delta _\mathcal {Z}$ and $\mathcal{Z} \times \mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ and use Lemma 101.7.7 and the fact that $\mathcal{Z} \times \mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ is separated. Thus we reduce to the case discussed in the next paragraph.

Assume $\mathcal{X}$ is quasi-compact, $\Delta $ is quasi-compact, $\mathcal{X}$ is reduced, and $|\mathcal{X}|$ irreducible. We have to show $|\mathcal{X}|$ has a unique generic point. Since $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is a base change of $\Delta $, we see that $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact (Lemma 101.7.3). Thus there exists a dense open substack $\mathcal{U} \subset \mathcal{X}$ which is a gerbe by Proposition 101.29.1. In other words, $|\mathcal{U}| \subset |\mathcal{X}|$ is open dense. Thus we may assume that $\mathcal{X}$ is a gerbe. Say $\mathcal{X} \to X$ turns $\mathcal{X}$ into a gerbe over the algebraic space $X$. Then $|\mathcal{X}| \cong |X|$ by Lemma 101.28.13. In particular, $X$ is quasi-compact. By Lemma 101.28.14 we see that $X$ has quasi-compact diagonal, i.e., $X$ is a quasi-separated algebraic space. Then $|X|$ is spectral by Properties of Spaces, Lemma 66.15.2 which implies what we want is true. $\square$