The Stacks project

Proof. Since $\mathcal X$ is algebraic, there exists a scheme $U$ with a surjective smooth morphism $U \to \mathcal X$. Let $U_ i \subset U$ be the open subscheme representing $\mathcal{X}_ i \times _{\mathcal X} U$ and $U' \subset U$ the open subscheme representing $\mathcal{X}' \times _{\mathcal X} U$. By hypothesis, $U'\subset \bigcup _{i\in I} U_ i$. From the proof of Lemma 100.6.2, there is a quasi-compact open $V \subset U'$ such that $V \to \mathcal{X}'$ is a surjective smooth morphism. Therefore there exists a finite subset $I' \subset I$ such that $V \subset \bigcup _{i \in I'} U_ i$. We claim that $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$. Take $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}'_ T)$ for $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Since $\mathcal{X}' \to \mathcal{X}$ is a monomorphism, we have cartesian squares

By base change, $V \times _{\mathcal X} T \to T$ is surjective. Therefore $\bigcup _{i \in I'} U_ i \times _{\mathcal X} T \to T$ is also surjective. Let $T_ i \subset T$ be the open subscheme representing $\mathcal{X}_ i \times _{\mathcal X} T$. By a formal argument, we have a Cartesian square

where the vertical arrows are surjective by base change. Since $U_ i \times _{\mathcal{X}_ i} T_ i \simeq U_ i \times _{\mathcal X} T$, we find that $\bigcup _{i \in I'} T_ i = T$. Hence $x$ is an object of $(\bigcup _{i\in I'} \mathcal{X}_ i)_ T$ by definition of the union. Observe that the inclusion $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$ is automatically an open substack. $\square$