 Solution

If $M=0$, all its localizations are zero. Conversely, take $0\ne m\in M$. Its annihilator $I={\mathrm{Ann}}_R(m)$ is a proper ideal, so it is contained in a maximal ideal $P$, which is prime. The zero criterion for module localization says that $m/1=0$ in $M_P$ only if some $s\notin P$ satisfies $sm=0$. This would put $s\in I\subseteq P$, a contradiction. Therefore $M_P\ne 0$, provingIn fact maximal ideals suffice, and neither finite generation nor Noetherianity is needed for this implication.