Let $S_g$ be the fundamental group of a genus $g$ orientable surface. Suppose that $N_1$ and $N_2$ are two infinite index nontrivial normal subgroups in $S_g$ (hence they are free of infinite rank).
Is it necessarily true that $N_1 \cap N_2$ is nontrivial?
Yes. We have $[N_1,N_2]\subset N_1\cap N_2$, and actually $[N_1,N_2]\neq 1$ as soon as $N_1\neq 1\neq N_2$ (infinite index is not the point).
Indeed, $S_g$ ($g\ge 2$) can be embedded in a Zariski-dense way in the simple complex group $PGL_2$. If $M_i$ is the Zariski closure of $N_i$, then $M_i$ is normal in $PGL_2$ and $[M_1,M_2]=1$, which implies that either $M_1$ or $M_2$ is reduced to 1, hence either $N_1$ or $N_2$ is reduced to 1.
This is one approach, there are many others.