Proper closed subsets of a curve are finite

Since $X$ is of finite type over $k$ it is locally of finite type and quasi-compact [F1], so the affine opens $U=\mathrm{Spec}A$ of $X$ with $k\to A$ of finite type form an open cover of $X$ by [F1] and contain a finite subcover $U_1=\mathrm{Spec}{A}_1,\dots ,U_r=\mathrm{Spec}{A}_r$. Each $A_i$ is a finite type $k$-algebra, hence a Noetherian ring by [F2] with $k$ a field, so each $U_i$ is a Noetherian topological space by [F3]. A finite union of Noetherian subspaces is Noetherian: for a descending chain $F_0\supseteq F_1\supseteq \cdots$ of closed subsets of $X$ the traces $F_m\cap U_i$ form a descending chain of closed subsets of $U_i$, which stabilizes from some index $m_i$ on, and then $F_m$ is constant for $m\ge {\max }_i{m}_i$ because the finitely many $U_i$ cover $X$. Hence the underlying space of $X$ is Noetherian in the sense of [F4].

Distinct points of $X$ have distinct closures. Suppose $x\mathrm{\ne }y$ with $\overline{\{x\}}=\overline{\{y\}}$. By [F5] choose an affine open $U=\mathrm{Spec}A$ containing $x$. For an open $V\subseteq X$ and a subset $S\subseteq X$ one has $\bar{S}\cap V={\overline{S\cap V}}^V$, where the second closure is taken in $V$: the inclusion $\supseteq$ is clear, and a point of $\bar{S}\cap V$ has every open neighbourhood in $V$ meeting $S$. Since $x\in \overline{\{y\}}$, this gives $x\in {\overline{\{y\}\cap U}}^U$; were $y\mathrm{\notin }U$ the right side would be empty, so $y\in U$. Writing ${\mathfrak{p}}_x,{\mathfrak{p}}_y$ for the primes of $A$ corresponding to $x,y$, the two membership statements $x\in \overline{\{y\}}$ and $y\in \overline{\{x\}}$ translate by [F6] into ${\mathfrak{p}}_x\supseteq {\mathfrak{p}}_y$ and ${\mathfrak{p}}_y\supseteq {\mathfrak{p}}_x$, so ${\mathfrak{p}}_x={\mathfrak{p}}_y$ and $x=y$, a contradiction.

By [F10] the scheme $X$ is nonempty and irreducible, and ${\eta }_X$ is its generic point by hypothesis; if $x$ is any generic point of $X$ then $\overline{\{x\}}=X=\overline{\{{\eta }_X\}}$, so $x={\eta }_X$ by step 1.2. Thus $X$ has a unique generic point. The same argument, applied to the closed subset $\overline{\{x\}}$, shows that every irreducible closed subset of $X$ has at most one generic point in the sense of [F12].

Let $Z⊊X$ be closed and nonempty. By step 1.1 and [F7] the subspace $Z$ is Noetherian, so by [F8] it is a finite union $Z=W_1\cup \cdots \cup W_m$ of nonempty irreducible closed subsets of $Z$. Each $W_j$ is closed in $X$ because $Z$ is closed in $X$, and $W_j⊊X$ because $W_j\subseteq Z⊊X$.

Each $W_j$ has chain dimension $0$. Otherwise $\dim W_j\ge 1$, so by [F9] there are nonempty irreducible closed subsets $Z_0⊊Z_1$ of $W_j$; these are closed in $X$ because $W_j$ is closed in $X$, and $Z_1⊊X$ by step 2.2, so $Z_0⊊Z_1⊊X$ is a strict chain of length two of nonempty irreducible closed subsets of $X$, contradicting $\dim X=1$.

Each $W_j$ is a single point. For $w\in W_j$, the closure of $\{w\}$ in $W_j$ is a nonempty irreducible closed subset by [F13]; if it were proper, it and $W_j$ would form a strict chain of length one of nonempty irreducible closed subsets, contrary to $\dim W_j=0$ by [F9]. Hence every point of $W_j$ is generic for $W_j$. Since $W_j$ is closed in $X$, any two such points have the same closure in $X$ and therefore coincide by step 1.2. Thus $W_j=\{w_j\}$ for a single point $w_j$.

By steps 2.2 and 4.1 the proper closed subset $Z$ is the finite set $\{w_1,\dots ,w_m\}$, and each $w_j$ is a closed point of $X$ because $W_j=\{w_j\}$ is closed in $X$ by step 2.2. For $Z=\mathrm{\varnothing }$ the same conclusion is vacuous, and $m=0$ is allowed in the displayed union. This proves claim 1.

Let $x\mathrm{\ne }{\eta }_X$. If $\overline{\{x\}}=X$ then $x$ is a generic point of $X$, so $x={\eta }_X$ by step 2.1; hence $\overline{\{x\}}$ is a proper closed subset of $X$, and step 5.1 applied to it shows that its points, in particular $x$, are closed points of $X$. This is claim 2, the uniqueness of the generic point having been shown in step 2.1.

The Axiom of Choice [F14] enters exactly where the cited suppliers assume it: at step 1.1 through [F3], at step 1.2 through [F6], and at step 2.2 through the minimal-counterexample argument in [F8]. The chain-dimension reasoning of steps 3.1–6.1 and the remaining topological arguments are choice-free. The empty proper closed subset, represented by $m=0$, is covered by step 5.1; dimensions $0$ and $\ge 2$ for $X$ are excluded by the hypothesis $\dim X=1$. ∎