Given an order ${\mathcal O}$ in a number field a regular prime ${\mathfrak p} \neq 0$ is defined by the condition ${\mathcal O}_{\mathfrak p}$ being integrally closed. (Neukirch : Algebraic Number Theory, Pg 79). I need to verify that there are only finitely many non-regular primes.
As I have not yet proved that the Picard group of ${\mathcal O}$ is finite, I don't see why this is immediate.
Here is the answer : As ${\mathcal O} \subseteq {\mathcal O}_K$ is a subring it embeds $\mathbb Z$ inside it. Hence the integral closure ${\tilde{\mathcal O}} \subseteq {\mathcal O}_K$. However ${\mathcal O}_K$ is the full integral closure of $\mathbb Z$ inside $K$. Hence ${\tilde{\mathcal O}} = {\mathcal O}_K$.
Once these two are equal, and the order contain a $\mathbb Z$-module of rank $n = [K : {\mathbb Q}]$, we have that ${\tilde{\mathcal O}}$ is a free $\mathbb Z$-module of rank $n$. Hence it must be a finitely generated ${\mathcal O}$-module.
Let ${\mathcal O}_K = {\mathcal O} \omega_1 + \dotsc + {\mathcal O} \omega_n$ where $\omega_1, \dotsc, \omega_n \in {\mathcal O}_K$. Now write $\omega_i = {\frac {a_i}{b_i}}$ for $a_i, b_i \in {\mathcal O}$ (as $K$ is the quotient field of $\mathcal O$). By definition, ${\mathfrak p} \leq {\mathcal O}$ is not regular if and only if $b_i \not\in {\mathfrak p}$ for some $i$. But ${\mathcal O}$ is Noetherian of dimension $1$, which means there are only finitely many primes $0 \neq b_i \not\in {\mathfrak p}$.