The Stacks project

Proof. Set $S_ n = \mathop{\mathrm{Spec}}(A_ n)$. Let $T_0 \subset T$ be the common image of the factorizations $S_ n \to T$. Then $T_0$ is quasi-compact. Let $T' \subset T$ be a quasi-compact open containing $T_0$. Then $S_ n \to T$ factors through $T'$. If we can show that $T' = S$, then $T' = T = S$. Hence we may assume $T$ is quasi-compact.

Assume $T$ is quasi-compact. In this case $T \to S$ is separated and quasi-finite (Morphisms, Lemma 29.20.15). Using Zariski's Main Theorem (in the form of More on Morphisms, Lemma 37.43.3) we choose a factorization $T \to W \to S$ with $W \to S$ finite and $T \to W$ an open immersion. Write $W = \mathop{\mathrm{Spec}}(B)$. The (unique) factorizations $S_ n \to T$ may be viewed as morphisms into $W$ and we obtain

Consider the morphism $h : S = \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B) = W$ coming from the arrow on the right. Then

is an open subscheme of $S$ containing the image of $S_ n \to S$ for all $n$. To finish the proof it suffices to show that any open $U \subset S$ containing the image of $S_ n \to S$ for some $n \geq 1$ is equal to $S$. This is true because $(A, \mathop{\mathrm{Ker}}(A \to A_ n))$ is a henselian pair (More on Algebra, Lemma 15.11.3) and hence every closed point of $S$ is contained in the image of $S_ n \to S$. $\square$