If $R$ is a ring, J is an ideal in $R$, and $I$ is an ideal of $J$ (with $J$ considered as a ring), does it follow that $I$ is an ideal of $R$? That is, is $I$ necessarily closed under multiplication by elements of $R$? Surely $I$ is closed under multiplication by elements of $J$ (since $I$ is an ideal of $J$). The "obvious" approach to proof fails so quickly that it must be false, that "an ideal of an ideal is not necessarily an ideal of the big ring". Please provide a counterexample. (Or a proof?)
Let $M$ be the matrix ring $M_2(\mathbb Q)$, and let $I=M$ viewed as simply a rational vector space. Consider the abelian group $R=M\oplus I$ and define on it a multiplication such that $$(a,v)\cdot(b,w)=(ab,aw+vb);$$ all products on the right hand side of this definition are good ol' matrix products.
You can easily check that this turns $R$ into a ring and that $I$ is an ideal of $R$ such that the product of any two elements of $I$ is zero in $R$. This has the immediate consequence that any $\mathbb Q$-subspace $J$ of $I$ is an ideal of $I$.
A little work will show, on the other hand, that $I$ does not properly contain any non-zero ideal of $R$.
N.B. I interpreted the word ideal in your question to mean bilateral ideal.