I am trying to prove this statement by deducing a contradiction to Krull's Principal ideal theorem. Here is my attempt:
Denote $R$ and $P$ the ring and the prime ideal under consideration, respectively, with $ht(P) \geq 2$. Assume there are finitely many prime ideals $\mathfrak{q}_1,...,\mathfrak{q}_n \subseteq P$ with $ht(\mathfrak{q}_i)=1$ for all $i=1,...n$. Consider the chain of prime ideals $\mathfrak{p}_i \subsetneq \mathfrak{q}_i \subsetneq{P}$ with $ht(\mathfrak{p}_i)=0$. For each $a \in P \setminus\mathfrak{q}_i$ there exist a chain of prime ideals $\mathfrak{p}_i \subsetneq \mathfrak{q}'_{a,i} \subsetneq{P}$ with $a \in \mathfrak{q}'_{a,i}$. Now we get the chain $\mathfrak{p}_i \subsetneq (a) \subseteq \mathfrak{q}'_{a,i} \subsetneq{P}$, and I don't know how to go on from this. I guess that $(a)=\mathfrak{q}'_{a,i}$ and then we can deduce a contradiction from Krull's Principal ideal theorem, but I am not sure if my intuition is right.
Any help or other approaches will be really appreciated.