I think the following works. Choose any non-zero element $f \in m_1 \cap m_2$ (if $m_1 \cap m_2 = 0$, then you weren't a domain). As pointed out in the comments, this doesn't work. Set $p$ to be a minimal associated prime of $f$. Somewhere in Matsumura it proves that this is height 1 I think.

EDIT: I was being dumb. Sorry about that, I shouldn't try to answer mathoverflow early in the morning. Anyway, if you could do it, then you need to argue by induction. But you need to choose your $f$ carefully.

EDIT: The following is junk. I'm not deleting it because of all the discussion in the comments.

I think the following works. Choose any non-zero element $f \in m_1 \cap m_2$ (if $m_1 \cap m_2 = 0$, then you weren't a domain). As pointed out in the comments, this doesn't work. Set $p$ to be a minimal associated prime of $f$. Somewhere in Matsumura it proves that this is height 1 I think.

EDIT: I was being dumb. Sorry about that, I shouldn't try to answer mathoverflow early in the morning. Anyway, if you could do it, then you need to argue by induction. But you need to choose your $f$ carefully.

I think the following works. Choose any non-zero element $f \in m_1 \cap m_2$ (if $m_1 \cap m_2 = 0$, then you weren't a domain). Set $p$ to be a minimal associated prime of $f$. Somewhere in Matsumura it proves that this is height 1 I think.

I think the following works. Choose any non-zero element $f \in m_1 \cap m_2$ (if $m_1 \cap m_2 = 0$, then you weren't a domain). As pointed out in the comments, this doesn't work. Set $p$ to be a minimal associated prime of $f$. Somewhere in Matsumura it proves that this is height 1 I think.

EDIT: I was being dumb. Sorry about that, I shouldn't try to answer mathoverflow early in the morning. Anyway, if you could do it, then you need to argue by induction. But you need to choose your $f$ carefully.