Finitely generated G-representation has "enough" simple quotients?
Solution 1:
Take $G=\langle a\rangle$ the infinite cyclic group, and $V=\mathbb{Q}^2$, with $a$ acting (on the right) through the matrix $\left(\begin{array}{cc}1&0\\1&1\end{array}\right)$. Your claim then is equivalent to the matrix being diagonalizable, which it is not. One can extend this example to infinite dimension by taking a direct sum with the regular module, which produces a counterexample to the claim for infinite-dimensional $V$.