Using Zorn's lemma is the approach I've seen. Would like feedback for a different approach I tried.
Similar/related question: Existence of minimal prime ideal contained in given prime ideal and containing a given subset
Proof?: Let $I$ be an ideal and $P$ be a prime ideal of the ring $R$ such that $I \subset P$. Let $$\mathscr{F}:=\{P'\subseteq R \vert (\text{P' is a prime ideal}) \land (I \subsetneq P' \subseteq P))\}$$ . Since $I \subset P \subseteq P$, then $P \in \mathscr{F}$. So $\mathscr{F}\neq \emptyset$. We assert that $\bigcap \mathscr{F} \neq \emptyset$ and that it's a prime ideal. Since $\forall P' \in \mathscr{F}(I \subsetneq P')$, then $I \subsetneq \bigcap\mathscr{F}$. Note $I \subsetneq \bigcap\mathscr{F}$ implies $\bigcap\mathscr{F}\neq \emptyset$. If $ab \in \bigcap\mathscr{F}$ and $a \notin \bigcap\mathscr{F}$, then $\forall P' \in \bigcap\mathscr{F}(ab \in P')$ and $a \notin P'$. Since $\forall P'\in\mathscr{F}$, $P'$ is a prime ideal, then $ab \in P'$ and $a \notin P'$ imply $b \in P'$, for all $P' \in \mathscr{F}$. Thus, $b \in \bigcap\mathscr{F}$, as desired. Lastly, we show $\bigcap\mathscr{F}$ is the minimal prime ideal of $I$. Suppose $\overline{P} \in \mathscr{F}$ such that $\overline{P} \subseteq \bigcap\mathscr{F}$. Since $\overline{P} \in \mathscr{F}$ imply $\bigcap\mathscr{F} \subseteq \overline{P}$, then $\bigcap\mathscr{F} = \overline{P}$. This equivalently, means there exist no prime ideal $\overline{P}$ where $I \subsetneq \overline{P} \subseteq \bigcap\mathscr{F}$, unless $\overline{P} = \bigcap\mathscr{F}$. Therefore, $\bigcap\mathscr{F}$ is the minimal prime ideal of $I$ contained in $P$.
