subideals of an ideal

Let us consider the sets$$R=\left\{ \begin{pmatrix} a & b & c\\ d & e & f\\ 0 & 0 & g \end{pmatrix}:a,b,c,d,e,f,g\in\mathbb{Z}\right\}\\J=\left\{\begin{pmatrix} 0&0&a\\0&0&b\\0&0&0\end{pmatrix}:a,b\in \mathbb{Z}\right\}\\I=\left\{\begin{pmatrix}0 &0&a\\0&0&0\\0&0&0\end{pmatrix}:a\in\mathbb{Z}\right\}$$Now we can easily check that $R$ is a ring with respect to usual matrix addition and multiplication and $J$ is an ideal of $R$ and $I$ is an ideal of $J$ but $I$ is not an ideal of $R$.

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.

I can provide a commutative counterexample: Let $R=\mathbb Q[X]/(X^2)$, and $I=(\bar X)$. Then the ideals of $R$ contained by $I$ are simply $I$ and $0$. But any additive subgroup of $I$ is an ideal of $I$ (since the multiplicative structure on $I$ is the trivial one).