Ideal of ideal needs not to be an ideal
Solution 1:
Consider $R=\mathbb Q[x]$, and $I=xR$ be the most obvious ideal of $R$.
Note that we can define $J$ as a subset of $I$ to be an ideal of $I$ if $J$ is a subgroup of $(I,+)$ and $IJ\subseteq J$. Find a $J$ that is a super-set of $x^2R$ but does not contain all of $I=xR$.