Solutions

By definition, there exist $N_1>0$ and $N_2<0$ such that for all $x$,

$$x>N_1\implies |f(x)|=f(x)<f(0)$$

and

$$x<N_2\implies |f(x)|=f(x)<f(0).$$

Consider the interval $D=[N_2,N_1]$. Since $f$ is continuous, there must exist $y\in D$ such that $f(y)\ge f(x)$ for all $x\in D$. But over this interval, $f(y)\ge f(0)$ since $0\in D$. Hence, $f(x)\le f(y)$ for all $x\in\mathbb{R}$.