= Solution
Write $X=\operatorname{Spec}B$, $S=\operatorname{Spec}A$ with $A,B$ <integral domains>, let $\alpha:A\to B$ be the <ring homomorphism>, and put $K=\operatorname{Frac}(B)$. We do not assume that $\alpha$ is injective. Write $A'=\alpha(A)\subseteq B\subseteq K$.
Let $D\subseteq K$ be any <valuation ring> containing $A'$, with <fraction field> $K$. The canonical <morphism of schemes> $\operatorname{Spec}K\to\operatorname{Spec}B$ and the <ring homomorphism> $A\to A'\hookrightarrow D$ 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 that
$$
B\xrightarrow{\beta}D\hookrightarrow K
$$
is 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> gives
$$
B\subseteq\bigcap_{\substack{D\subseteq K\text{ valuation ring}\\ A'\subseteq D}}D
=\overline{A'}^{\,K}.
$$
Thus every element of $B$ is an <integral element> over $A'$, and hence over $A$: lift the coefficients of its <monic polynomial> along the surjection $A\to A'$. 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 $B=A[b_1,\ldots,b_s]$. If $b_i$ satisfies a <monic polynomial> of degree $n_i$, repeatedly reducing its powers shows that the finitely many <monomials>
$$
b_1^{e_1}\cdots b_s^{e_s},\qquad 0\leq e_i<n_i,
$$
generate $B$ as an $A$-<module>. Equivalently, apply the allowed equivalence between finite and integral plus <finite type>. Hence
$$
\boxed{f\text{ proper between affine integral schemes}\Longrightarrow f\text{ finite}.}
$$
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