**Proof.**
Because $S/R$ is quasi-finite we may apply Theorem 10.123.12 to each point $\mathfrak q$ of $\mathop{\mathrm{Spec}}(S)$. Since $\mathop{\mathrm{Spec}}(S)$ is quasi-compact, see Lemma 10.17.8, we may choose a finite number of $g_ i \in S'$, $i = 1, \ldots , n$ such that $S'_{g_ i} = S_{g_ i}$, and such that $g_1, \ldots , g_ n$ generate the unit ideal in $S$ (in other words the standard opens of $\mathop{\mathrm{Spec}}(S)$ associated to $g_1, \ldots , g_ n$ cover all of $\mathop{\mathrm{Spec}}(S)$).

Suppose that $D(g) \subset \mathop{\mathrm{Spec}}(S')$ is contained in the image. Then $D(g) \subset \bigcup D(g_ i)$. In other words, $g_1, \ldots , g_ n$ generate the unit ideal of $S'_ g$. Note that $S'_{gg_ i} \cong S_{gg_ i}$ by our choice of $g_ i$. Hence $S'_ g \cong S_ g$ by Lemma 10.23.2.

We construct a finite algebra $S'' \subset S'$ as in (3). To do this note that each $S'_{g_ i} \cong S_{g_ i}$ is a finite type $R$-algebra. For each $i$ pick some elements $y_{ij} \in S'$ such that each $S'_{g_ i}$ is generated as $R$-algebra by $1/g_ i$ and the elements $y_{ij}$. Then set $S''$ equal to the sub $R$-algebra of $S'$ generated by all $g_ i$ and all the $y_{ij}$. Details omitted.
$\square$

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