Prove of specific prime ideals between a strict inclusion of prime ideals 
Statement. Let $R$ be a commutative ring with prime ideals $\mathfrak{p}$ and $\mathfrak{q}$. If $\mathfrak{p}\subsetneq\mathfrak{q}$, then there are prime ideals $\mathfrak{p}_1$ and $\mathfrak{q}_1$ in $R$ such that $\mathfrak{p}\subseteq\mathfrak{p}_1\subsetneq\mathfrak{q}_1\subseteq\mathfrak{q}$ and there are no other prime ideals between $\mathfrak{p}_1$ and $\mathfrak{q}_1$.

Here are my attempts. Consider the quotient ring $R/\mathfrak{p}$, which is an integral domain now.
Besides, we have $\mathfrak{p}/\mathfrak{p}=\langle 0\rangle$ and $\mathfrak{q}/\mathfrak{p}\ne\langle 0\rangle$, both of which are prime ideals in $R/\mathfrak{p}$ as well.
Then how can we find such prime ideals $\mathfrak{p}_1$ and $\mathfrak{q}_1$? I was thinking of minimal prime ideals over $\mathfrak{p}$, but since $\mathfrak{p}$ itself is prime, it make no sense to consider it now. Any help will be appreciated.
 A: Consider the set $$S=\{(\mathfrak{p}',\mathfrak{q}')\in\operatorname{Spec}(R)\times\operatorname{Spec}(R)\mid\mathfrak{p}\subseteq\mathfrak{p}'\subsetneq\mathfrak{q}'\subseteq\mathfrak{q}\},$$ which is certainly nonempty as $(\mathfrak{p},\mathfrak{q})$ is contained in it. Next, define a partial order on $S$ by $$(\mathfrak{p}_1,\mathfrak{q}_1)\le(\mathfrak{p}_2,\mathfrak{q}_2)\iff\mathfrak{p}_1\subseteq\mathfrak{p}_2~\text{and}~\mathfrak{q}_2\subseteq\mathfrak{q}_1.$$ One can easily verify that $\le$ defines a partial order on $S$. Consider an ascending chain $(\mathfrak{p}_i,\mathfrak{q}_i)_{i\in\Gamma}$ in $S$. Define $$\mathfrak{p}_0:=\bigcup_{i\in\Gamma}{\mathfrak{p}_i}\quad\text{and}\quad\mathfrak{q}_0:=\bigcap_{i\in\Gamma}{\mathfrak{q}_i}.$$ Again, one can also check that both $\mathfrak{p}_0$ and $\mathfrak{q}_0$ are prime ideals in $R$ with $\mathfrak{p}_0\subsetneq\mathfrak{q}_0$, hence $(\mathfrak{p}_0,\mathfrak{q}_0)\in S$ as well. Finally, we simply let $(\mathfrak{p}_1,\mathfrak{q}_1)$ be the maximal element in $S$ ensured by Zorn's lemma.
