**Proof.**
Choose any valuation ring $A \subset K$ dominating $R$ (which exist by Lemma 10.50.2). Denote $v$ the corresponding valuation. Let $x_1, \ldots , x_ r$ be a minimal set of generators of the maximal ideal $\mathfrak m$ of $R$. We may and do assume that $v(x_ r) = \min \{ v(x_1), \ldots , v(x_ r)\} $. Consider the ring

\[ S = R[x_1/x_ r, x_2/x_ r, \ldots , x_{r - 1}/x_ r] \subset K. \]

Note that $\mathfrak mS = x_ rS$ is a principal ideal. Note that $S \subset A$ and that $v(x_ r) > 0$, hence we see that $x_ rS \not= S$. Choose a minimal prime $\mathfrak q$ over $x_ rS$. Then $\text{height}(\mathfrak q) = 1$ by Lemma 10.60.11 and $\mathfrak q$ lies over $\mathfrak m$. Hence we see that $R' = S_{\mathfrak q}$ is a solution.
$\square$

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