Solutions

Suppose that $f$ is continuous at $a$. Pick any $\varepsilon>0$. By definition, there exists $\delta>0$ such that

$$\begin{align*} |x-a|<\delta &\implies |f(x)-f(a)|<\frac{\varepsilon}{2}\\ &\implies -\frac{\varepsilon}{2}+f(a)<f(x)<\frac{\varepsilon}{2}+f(a)\\ &\implies -\frac{\varepsilon}{2}-f(a)<f(x)<\frac{\varepsilon}{2}-f(a). \end{align*}$$

Hence, it follows that for all $x$ and $y$,

$$|x-a|<\delta \land |y-a|<\delta \implies |f(x)-f(a)|<\frac{\varepsilon}{2}$$$$\land\quad |f(y)-f(a)|<\frac{\varepsilon}{2}$$$$\implies -\varepsilon<f(x)-f(y)<\varepsilon.$$