This should be easy, but apparantly not for me. Let G be a topological group, and let $\mathcal{N}$ be a neighbourhood base for the identity element $e$ of $G$. Then for all $N_1,N_2 \in \mathcal{N}$, there exists an $N' \in \mathcal{N}$ such that $e \in N'\subset N_1 \cap N_2.$
This means for example that every $N \in \mathcal{N}$ contains the identity element, which seems strange to me.
As has been pointed out in the comments, this has nothing to do with groups; this is purely about what the term "neighborhood basis" means.
Given a point x in a topological space X and a neighborhood basis $\mathcal{N}$ for x, and $N_1, N_2 \in \mathcal{N}$, since $N_1, N_2$ are in this neighborhood basis, they are, in particular, neighborhoods of x. (And, note, therefore contain x.) Thus so is their interesection. And any neighborhood of x must contain some element of $\mathcal{N}$ (which in turn contains x, as they all do), because that's what it means for $\mathcal{N}$ to be a neighborhood basis.
(I've been deliberately ambiguous as to whether "neighborhoods" here are required to be open or not, because the above works either way. :) )