 Solution

Write $X=\mathrm{Spec}B$, $S=\mathrm{Spec}A$ with $A,B$ integral domains, let $\alpha :A\to B$ be the ring homomorphism, and put $K=\mathrm{Frac}(B)$. We do not assume that $\alpha$ is injective. Write $A^{\prime }=\alpha (A)\subseteq B\subseteq K$.

Let $D\subseteq K$ be any valuation ring containing $A^{\prime }$, with fraction field $K$. The canonical morphism of schemes $\mathrm{Spec}K\to \mathrm{Spec}B$ and the ring homomorphism $A\to A^{\prime }\hookrightarrow D$ form a valuative problem for $f$. Since $f$ is a proper morphism, there is an extension. On coordinate rings it is a ring homomorphism $\beta :B\to D$ such thatis the original inclusion of $B$ into $K$. Therefore every element of $B$ lies in $D$. Since $D$ was arbitrary, the stated description of integral closure as an intersection of valuation rings givesThus every element of $B$ is an integral element over $A^{\prime }$, and hence over $A$: lift the coefficients of its monic polynomial along the surjection $A\to A^{\prime }$. The ring homomorphism $\alpha$ is therefore an integral extension in the sense of the question.

A proper morphism is of finite type, so in this affine situation $B=A[b_1,\dots ,b_s]$. If $b_i$ satisfies a monic polynomial of degree $n_i$, repeatedly reducing its powers shows that the finitely many monomialsgenerate $B$ as an $A$-module. Equivalently, apply the allowed equivalence between finite and integral plus finite type. Hence