My idea: Let $f \equiv 1$ and set define $f_{n}\equiv 1_{[-n,n]}$ clearly then for any $N \in \mathbb N$ and $\epsilon>0$ there is $n_{0}$ large enough such that $\forall n \geq n_{0}$ it follows that $\vert \vert f_{n}-f\vert \vert_{L^{\infty}([-N,N])}=0$ but equally on $\mathbb R$:
$\vert \vert f_{n}-f\vert \vert_{L^{\infty}(\mathbb R)}=1$ any $n \in \mathbb N$. Is my idea correct?