Lemma 1 Let $A$ be a discrete valuation ring and $K$ be its field of fractions. Let $p$ be its maximal ideal. Let $B$ be a subring of $K$ such that $A \subset B$. Suppose $B$ is a local ring and $p = q \cap A$, where $q$ is its maximal ideal.
Then $A = B$.
Proof: Suppose $A \neq B$. There exists $b \in B - A$. Then $\frac{1}{b} \in p$. Hence $1/b \in q$. Since $b \in B$, $1 = b\frac{1}{b} \in q$. This is a contradiction. QED
Lemma 2 Let $K$ be a finite extension field of $\mathbb{Q}$, $\mathcal{O}$ the integral closure of $\mathbb{Z}$ in $K$. Let $R$ be a discrete valuation ring of $K$. Then there exists a nonzero prime ideal $p$ of $\mathcal{O}$ such that $R = \mathcal{O}_p$.
Proof: Let $m$ be the maximal ideal of $R$. Since $1 \in R$, $\mathbb{Z} \subset R$. Let $\alpha \in \mathcal{O}$. Since $\alpha$ is integral over $\mathbb{Z}$, it is integral over $R$. Since $R$ is a unique factorization domain, $R$ is integrally closed in $K$. Hence $\alpha \in R$. Hence $\mathcal{O} \subset R$. Let $p = m \cap \mathcal{O}$. Then $\mathcal{O}_p \subset R_m = R$. Suppose $p = 0$. Then $\mathcal{O}_p = K \subset R$. This is a contradiction. Hence $\mathcal{O}_p$ is a discrete valuation ring. Clearly $p\mathcal{O}_p = m \cap \mathcal{O}_p$. Hence $\mathcal{O}_p = R$ by Lemma 1. QED
Lemma 3. Let $A$ be an integral domain, $K$ its field of fractions. Then $A = \bigcap_m A_m$, where $m$ runs through all maximal ideals of $A$.
Proof: Clearly $A \subset \bigcap_m A_m$. Hence it suffices to prove the other inclusion.
Let $x \in \bigcap_m A_m$. Suppose $x$ does not belong to $A$. Let $I = \{a \in A\colon ax \in A\}$. Clearly $I$ is an ideal of $A$. Since $x$ does not belong to $A$, $I \neq A$. Hence there exista maximal ideal $m$ such that $I \subset m$. Since $x \in A_m$, there exis $s \in A - m$ such thjat $sx = 0$. This is a contradiction. Hence $x \in A$. Hence $\bigcap_m A_m \subset A$. QED
Theorem Let $K$ be a finite extension field of $\mathbb{Q}$. Let $\mathcal{O}$ be the integral closure of $\mathbb{Z}$ in $K$. Let $R$ be a subring of $K$ such that $K$ is its field of fractions. Suppose $R$ is Dedekind. Then there exists a set $T$ of nonzero prime ideals of $\mathcal{O}$ such that $R = \bigcap_{p\in T} \mathcal{O}_p$.
Proof: Let $m$ be a nonzero prime ideal of $R$. Since $R$ is Dedekind, $R_m$ is a discrete valuation ring. Hence the assertion follows immediately from Lemma 2 and Lemma 3. QED