Let $R$ be a commutative ring with unity and $I \subsetneq R$ an ideal that is maximal among the proper subideals of $R$. Let $K \subsetneq I$ be an ideal. Must it necessarily be true that there exists an ideal $J$ with $K \subseteq J \subsetneq I$ which is maximal among the proper subideals of $I$?
For example, this is true if $I$ is finitely generated: Assume that $I = \langle x_1, \ldots, x_n \rangle$. Let $\mathcal{A}$ be a chain of ideals such that for each $A \in \mathcal{A}$, we have $K \subseteq A \subsetneq I$. Let $B = \bigcup \mathcal{A}$. Then $B$ is an ideal with $K \subseteq B$. We also have $B \subseteq I$. Striving for a contradiction, assume that $B = I$. Now, for each $1 \leq k \leq n$, we have $x_k \in B$, so there must exist $A_k \in \mathcal{A}$ with $x_k \in A_k$. Because $\mathcal{A}$ is totally ordered with respect to inclusion, we can let $A = \bigcup_{k=1}^n A_k \in \mathcal{A}$. But now $A$ is an ideal with $x_1, \ldots, x_n \in A$, so we have $I = \langle x_1, \ldots, x_n \rangle \subseteq A$. But this would imply $A = I$, a contradiction. Hence we must have $B \subsetneq I$ and therefore $K \subseteq B \subsetneq I$. Therefore, every chain $\mathcal{A}$ of ideals $A$ with $K \subseteq A \subsetneq I$ has an upper bound $B$ with $K \subseteq B \subsetneq I$, and hence by Zorn's Lemma, we have a maximal ideal $J$ with $K \subseteq J \subsetneq I$.
Best Answer
By passing to the quotient ring $R/K$ we see it suffices to solve the problem when $K=0$, so we seek a maximal subideal of $0 \neq I' = I/K$. Viewed as an $R$-module, your proof has rediscovered that finitely generated modules contain maximal submodules (although note that you require $I \neq 0$).
Here is a counterexample. Let $R = \mathbb{Q}[x_1, x_2, \dots]/(x_1^2, x_2^2 - x_1, x_3^2 - x_2, \dots)$ with maximal ideal $I = (x_1, x_2, \dots)$. Suppose $J$ is a maximal subideal. Note if $x_n \in J$ then $x_{n-1}, \dots, x_1 \in J$. Take the smallest $n$ such that $x_n \not\in J$. So $x_1, \ldots, x_{n-1} \in J$ and $x_n \not\in J$ and so $I = J + (x_n)$. So $x_{n+1} = j + rx_n$ and hence $x_{n+1} - rx_n \in J$. By difference of two squares, $x_n - r^2 x_{n-1} = x_{n+1}^2 - r^2 x_n^2 = (x_{n+1} - r x_n)(x_{n+1} + r x_n) \in J$ and as $x_{n-1} \in J$ we get $x_n \in J$, a contradiction.