Here is the proof that the sum of two functions is continuous.
Let f and g be continuous at some x. Let $\epsilon>0$ be chosen arbitrarily. Since f is continuous, there exists $\delta_f>0$ such that if $|x-u|<\delta_f$ then $|f(x)-f(u)|<\frac{\epsilon}{2}$. Similarly, choose $\delta_g$ such that if $|x-u|<\delta_g$ then $|g(x)-g(u)|<\frac{\epsilon}{2}$.
Now set $\delta=\min\{\delta_f,\delta_g\}$. Then for any u with $|x-u|<\delta$, both $|x-u|<\delta_f$ and $|x-u|<\delta_g$ and so
(1)
\begin{align} |f(x)+g(x)-(f(u)+g(u))|=|f(x)-f(u)+g(x)-g(u)| \leq |f(x)-f(u)| + |g(x)-g(u)| < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \quad \blacksquare \end{align}