Let $S$ be a set, $G$ its free group and $\mathcal{N}$ the set of normal $N \leq G$ such that $[G:N] < \infty$. Prove that $$ \bigcap_{N \in \mathcal{N}} N \ = \ \{ e\} $$
I know how free groups are constructed and defined, aside from that I know very little about them. I don't have much knowledge about covering spaces either, even though that theory is frequently used when dealing with free groups.
Given $x \in G \setminus \{e\}$, is there any straight forward way to construct a normal subgroup $N$ of finite index such that $x \notin N$? I know that this $N$ has to be free as well, but I really don't know what generators I have to choose. I hope you can help me.