I propose the following: suppose $\,U\,$ is not prime, thus there exist $$\,x,y\in R \text{ such that }x,y\notin U\,,\,xy\in U\,.$$ Define now $B:=U+\langle y\rangle$.
By maximality of $\,U\,$ we have that $\,B\,$ is f.g., say $$\,B=\Bigl\langle u_i+r_iy\,,\,1\leq i\leq k\,,\,k\in\mathbb{N}\,\,;\,\,u_i\in U\,,\,r_i\in R\Bigr\rangle$$ and let now $$U_y:=\{s\in R\,\,;\,\,sy\in U\}.$$ (1) Check that $\,U_y\,$ is a proper ideal in $\,R\,$.
(2) Show that $\,U_y\,$ is f.g.
Put $\,U_y=\langle s_1,\ldots,s_m\rangle\,$, and take $\,u\in U\Longrightarrow\,\exists v_1,\ldots,v_k, t_1,\ldots,t_k\in R\,\,s.t.$$$u=\sum_{n=1}^kv_nu_n+\sum_{n=1}^kt_nr_ny.$$
(3) Show that $\displaystyle{\sum_{n=1}^kt_nr_n}\in U_y.$
(4) Putting $\,\Omega:=\{u_i,\ldots,u_k,ys_1,\ldots,ys_m\}\,$, derive the contradiction $\,U=\langle\Omega\rangle$.
Hint: The intersection of a chain of prime ideals is prime.
Full solution:
The idea is to apply Zorn's lemma downwards. Consider the set $S:=\{Q \ | \ Q\text{ is a prime ideal of } R, X\subseteq Q\subseteq P\}$ ordered by inclusion. Note that $P\in S$ implies $S\neq\emptyset$. Consider a chain $\{Q_i\}_{i\in I}$ in $S$. Then $\bigcap_{i\in I} Q_i$ is in $S$, so by Zorn's lemma there exists an element minimal in $S$, i.e., minimal with respect being prime inside $P$ and containing $X$.
Let us see that $Q:=\bigcap_{i\in I} Q_i$ is actually prime: let $ab\in Q$ with $a,b\in R\setminus Q$. Then there exist indices $j,k$ such that $a\not\in Q_j$, $b\not\in Q_k$, say with $Q_j\supseteq Q_k$. Therefore $a,b\not\in Q_k$, but $ab\in Q$ implies $ab\in Q_k$, contradicting the primeness of $Q_k$.
Regarding your last question: Just apply the above proposition with $X:=0$ to get a minimal prime ideal of $R$ inside $P$.
Best Answer
Suppose $I\nsubseteq P$, say $i\in I\setminus P$. Let $j\in J$. Since $ij\in P$ and $P$ is prime, either $i\in P$ or $j\in P$. But $i\notin P$, hence $j\in P$.
$j\in J$ was arbitraty, so $J\subseteq P$.