Solutions

Pick $\varepsilon=|\alpha|\ne 0$. By definition, there exists $\delta>0$ such that

$$\begin{align*} |x-a|<\delta &\implies |f(x)|<|\alpha|\\ &\implies -|\alpha|<f(x)<|\alpha|\\ &\implies -|\alpha|+\alpha<f(x)+\alpha<|\alpha|+\alpha. \end{align*}$$

and it follows that $f+\alpha$ is non-zero regardless of the sign of $\alpha$.