Can we find a submodule $I$ such that it contains no irreducible submodules of $M$, yet it is not the zero module?
If $M$ is semisimple, we know that every submodule must contain at least one irreducible sub-submodule, but does it hold if $M$ is not semisimple?
No submodule of the $\mathbb Z$-module $\mathbb Z$ contains any simple $\mathbb Z$-submodules.
There is a name for the class of rings for which you can't find examples. A ring is said to be right semi-Artinian if every nonzero module over the ring has a nonzero socle. That means every nonzero module contains a simple submodule, and since submodules of modules are still modules, that would preclude counterexamples from existing in a right semi-Artinian ring.
The question in general about whether an $R$ module $M$ contains a simple submodule can be summed up as "it contains a simple submodule if and only if its socle is nonzero." Whether or not this is the case depends entirely on the module's structure, and it may be the case, or it may not be the case.