Non-finitely generated R-module

Solution 1:

A (commutative with identity) ring $R$ is Noetherian if and only if every ideal of $R$ is finitely generated (as a module over $R$ or as an ideal, it's the same).

A ring $R$ is Noetherian if and only if every submodule of every finitely generated $R$-module is finitely generated.

Thus your example requires a ring that's not Noetherian and the simplest example is $R=\mathbb{Z}[x_1,x_2,\dotsc]$, the ring of polynomials on countably many indeterminates over $\mathbb{Z}$, because the ideal $I$ generated by the indeterminates isn't finitely generated.

Indeed there is the strictly increasing sequence of ideal $$ (0)\subset (x_1)\subset (x_1,x_2)\subset\dotsb $$ and the union of these ideals is $I$. Why is this sequence strictly increasing? Because if $$ x_{n+1}=x_1f_1+x_2f_2+\dots+x_nf_n \qquad(f_1,f_2,\dots,f_n\in R) $$ evaluating at $x_1=x_2=\dots=x_n=0$ and $x_{n+1}=1$ leads to a contradiction. But if $I$ is finitely generated, its generators belong to one of the ideals in the chain: again a contradiction.

Note that we're talking about finitely generated as modules, not as rings.