As can be seen from the comment, the theorem and the proof are wrong. I'm leaving this for posterity & in case someone wants to find the mistake.
Proof: Let $\mathfrak{m}=R\backslash R^\times$ be the maximal ideal of $R$, and let $\mathfrak{m}=tR$.
Let $t=a/b$ for some $a,b\in D$. Since $R\subsetneq K$, $t\neq 0$ so $a,b\neq 0$. Since $b\in D$, $b$ is invertible in $R$ so we can choose a different uniformizing parameter, to be $t_0=a$.
Let $t_0=x_0y_0$ with $x_0,y_0\in D$. Then $x_0,y_0\in R$ so WLOG $x_0$ is irreducible in $R$ and $y_0$ is invertible in $R$. Let $t_1=x_0y_0y_0^{-1}=x_0$ be a different uniformizing parameter in $R$. Then $t_1 \mid t_0$ and $t_0D\subseteq t_1D$, we can continue to make a chain of ideals like this $t_0D\subseteq t_1D\subseteq \dots t_iD \dots$ Since $D$ is Noetherian, and thus there must be some $i$ for which $t_{i+1}=t_{i}$ so $t_i$ is irreducible in $D$. Let $t=t_i$ be our uniformizing parameter for $\mathfrak{m}$.
Let $\mathfrak{p}=\mathfrak{m}\cap D$ be an ideal in $D$. From the work we did before we know that $t\in D$ so $\{0\}\neq\mathfrak{p}=tD$ and $t$ is irreducible so $tD$ is a non-0 prime ideal.
Let $R'=D_\mathfrak{p}$. Since $D$ is a noetherian domain, $R'$ is a noetherian local domain that is not a field, so it is a DVR. The maximal ideal of $R'$ is $tR'=tD_\mathfrak{p}$. Let $R'' = R'\cap R$. Since $R''$ is the intersection of two DVR's, it is a DVR. Since $tR''=tR'\cap tR$ is its maximal ideal, we have: $D\subseteq R'' \subseteq R \subsetneq K$ and $D\subseteq R''\subseteq R\subsetneq K$ and $tR''\subseteq tR$ and $tR''\subseteq tR'$. Thus from Problem 2.26b for both cases we get $R'=R''=R$ so $R=D_\mathfrak{p}$ as required.
For reference, here is Problem 2.26:
Let $R$ be a DVR with quotient field $K$; Let $\mathfrak{m}$ be the maximal ideal of $R$.
- Show that if $z\in K$, $z\notin R$, then $z^{-1}\in \mathfrak{m}$.
- Suppose $R\subseteq S\subseteq K$, and $S$ is also a DVR. Suppose the maximal ideal of $S$ contains $\mathfrak{m}$. Show that $S=R$.
Proof:
- Let $z=a/b$ for $a,b \in R$. $z\notin R \Rightarrow b\notin R^\times$ so $b\in \mathfrak{m}$. If $a\in R^\times$ then $a^{-1}\in R \Rightarrow z^{-1}=b/a=ba^{-1}\in \mathfrak{m}$. If $a\notin R^\times$ then $a\in \mathfrak{m}$. Let $a=ut^n$ and $b=vt^m$ for some uniformizing parameter $t$. Since $z\notin R$ we have $m > n$. Thus $b/a=(v/u)t^{m-n}\in \mathfrak{m}$.
- Let $\mathfrak{n}$ be the maximal ideal of $S$. Assume BWOC that there is some element $z\in S\backslash R$. Then this $z$ satisfies the conditions of $1$, so $z^{-1}\in \mathfrak{m}\Rightarrow z^{-1}\in \mathfrak{n}$. But we got that on the one hand both $z,\ z^{-1}\in S$ so $z^{-1}$ is invertible, but at the same time $z^{-1}\in \mathfrak{n}$ so it is not invertible. Which is a contradiction.
