The Stacks project

Proof. After replacing $U_ n$ by $\bigcap _{i = 1, \ldots , n} U_ i$ we may assume that $U_1 \supset U_2 \supset \ldots$. Let $x \in X$. We will show that $x$ is in the closure of $\bigcap _{n \geq 1} U_ n$. Thus let $E$ be a neighbourhood of $x$. To show that $E \cap \bigcap _{n \geq 1} U_ n$ is nonempty we may replace $E$ by a smaller neighbourhood. After replacing $E$ by a smaller neighbourhood, we may assume that $E$ is quasi-compact.

Set $x_0 = x$ and $E_0 = E$. Below, we will inductively choose a point $x_ i \in E_{i - 1} \cap U_ i$ and a quasi-compact neighbourhood $E_ i$ of $x_ i$ with $E_ i \subset E_{i - 1} \cap U_ i$. Because $X$ is Hausdorff, the subsets $E_ i \subset X$ are closed (Lemma 5.12.4). Since the $E_ i$ are also nonempty we conclude that $\bigcap _{i \geq 1} E_ i$ is nonempty (Lemma 5.12.6). Since $\bigcap _{i \geq 1} E_ i \subset E \cap \bigcap _{n \geq 1} U_ n$ this proves the lemma.

The base case $i = 0$ we have done above. Induction step. Since $E_{i - 1}$ is a neighbourhood of $x_{i - 1}$ we can find an open $x_{i - 1} \in W \subset E_{i - 1}$. Since $U_ i$ is dense in $X$ we see that $W \cap U_ i$ is nonempty. Pick any $x_ i \in W \cap U_ i$. By definition of locally quasi-compact spaces we can find a quasi-compact neighbourhood $E_ i$ of $x_ i$ contained in $W \cap U_ i$. Then $E_ i \subset E_{i - 1} \cap U_ i$ as desired. $\square$