The Stacks project

Proof. Let $x, y \in X$ with $x \not= y$. Since $X$ is sober, there is an open subset $U$ containing exactly one of the two points $x, y$. Say $x \in U$. We may replace $U$ by a quasi-compact open neighbourhood of $x$ contained in $U$. Then $U$ and $U^ c$ are open and closed in the constructible topology. Hence $X$ is Hausdorff in the constructible topology because $x \in U$ and $y \in U^ c$ are disjoint opens in the constructible topology. The existence of $U$ also implies $x$ and $y$ are in distinct connected components in the constructible topology, whence $X$ is totally disconnected in the constructible topology.

Let $\mathcal{B}$ be the collection of subsets $B \subset X$ with $B$ either quasi-compact open or closed with quasi-compact complement. If $B \in \mathcal{B}$ then $B^ c \in \mathcal{B}$. It suffices to show every covering $X = \bigcup _{i \in I} B_ i$ with $B_ i \in \mathcal{B}$ has a finite refinement, see Lemma 5.12.15. Taking complements we see that we have to show that any family $\{ B_ i\} _{i \in I}$ of elements of $\mathcal{B}$ such that $B_{i_1} \cap \ldots \cap B_{i_ n} \not= \emptyset $ for all $n$ and all $i_1, \ldots , i_ n \in I$ has a common point of intersection. We may and do assume $B_ i \not= B_{i'}$ for $i \not= i'$.

To get a contradiction assume $\{ B_ i\} _{i \in I}$ is a family of elements of $\mathcal{B}$ having the finite intersection property but empty intersection. An application of Zorn's lemma shows that we may assume our family is maximal (details omitted). Let $I' \subset I$ be those indices such that $B_ i$ is closed and set $Z = \bigcap _{i \in I'} B_ i$. This is a closed subset of $X$ which is nonempty by Lemma 5.12.6. If $Z$ is reducible, then we can write $Z = Z' \cup Z''$ as a union of two closed subsets, neither equal to $Z$. This means in particular that we can find a quasi-compact open $U' \subset X$ meeting $Z'$ but not $Z''$. Similarly, we can find a quasi-compact open $U'' \subset X$ meeting $Z''$ but not $Z'$. Set $B' = X \setminus U'$ and $B'' = X \setminus U''$. Note that $Z'' \subset B'$ and $Z' \subset B''$. If there exist a finite number of indices $i_1, \ldots , i_ n \in I$ such that $B' \cap B_{i_1} \cap \ldots \cap B_{i_ n} = \emptyset $ as well as a finite number of indices $j_1, \ldots , j_ m \in I$ such that $B'' \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset $ then we find that $Z \cap B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset $. However, the set $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m}$ is quasi-compact hence we would find a finite number of indices $i'_1, \ldots , i'_ l \in I'$ with $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} \cap B_{i'_1} \cap \ldots \cap B_{i'_ l} = \emptyset $, a contradiction. Thus we see that we may add either $B'$ or $B''$ to the given family contradicting maximality. We conclude that $Z$ is irreducible. However, this leads to a contradiction as well, as now every nonempty (by the same argument as above) open $Z \cap B_ i$ for $i \in I \setminus I'$ contains the unique generic point of $Z$. This contradiction proves the lemma. $\square$