**Proof.**
There is an immediate reduction to the case $S = \mathop{\mathrm{Spec}}(R)$ where $R$ is a Nagata ring by assumption (1). We have to show that the integral closure $A$ of $R$ in $\Gamma (X, \mathcal{O}_ X)$ is finite over $R$. Since $f$ is quasi-compact by assumption (2) we can write $X = \bigcup _{i = 1, \ldots , n} U_ i$ with each $U_ i$ affine. Say $U_ i = \mathop{\mathrm{Spec}}(B_ i)$. Each $B_ i$ is reduced by assumption (5) and has finitely many minimal primes $\mathfrak q_{i1}, \ldots , \mathfrak q_{im_ i}$ by assumption (3) and Algebra, Lemma 10.26.1. We have

\[ \Gamma (X, \mathcal{O}_ X) \subset B_1 \times \ldots \times B_ n \subset \prod \nolimits _{i = 1, \ldots , n} \prod \nolimits _{j = 1, \ldots , m_ i} (B_ i)_{\mathfrak q_{ij}} \]

the second inclusion by Algebra, Lemma 10.25.2. We have $\kappa (\mathfrak q_{ij}) = (B_ i)_{\mathfrak q_{ij}}$ by Algebra, Lemma 10.25.1. Hence the integral closure $A$ of $R$ in $\Gamma (X, \mathcal{O}_ X)$ is contained in the product of the integral closures $A_{ij}$ of $R$ in $\kappa (\mathfrak q_{ij})$. Since $R$ is Noetherian it suffices to show that $A_{ij}$ is a finite $R$-module for each $i, j$. Let $\mathfrak p_{ij} \subset R$ be the image of $\mathfrak q_{ij}$. As $\kappa (\mathfrak q_{ij})/\kappa (\mathfrak p_{ij})$ is a finitely generated field extension by assumption (4), we see that $R \to \kappa (\mathfrak q_{ij})$ is essentially of finite type. Thus $R \to A_{ij}$ is finite by Algebra, Lemma 10.162.2.
$\square$

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