Theorem: Let $f_n$ continuous functions defined on a set $A\subseteq\mathbb{R}$, and assume $\sum_{n=1}^{\infty}f_n$ converges uniformly on $A$ to a function $f$. Then $f$ is continuous on $A$.
My Approach: Let $\epsilon>0$ and fix $c\in A$. Choose $N$ so that $\forall x\in A$ $$\underbrace{\left|f(x)-\sum_{n=1}^{\infty}f_N(c)\right|}_{\text{using uniform convergence}}<\frac{\epsilon}{3}$$ Because $f_N$ is continuous, there exists a $\delta>0$ for which $$|f_N(x)-f_n(c)|<\frac{\epsilon}{3n}$$ is true whenever $|x-c|<\delta$. But this implies,
$$\begin{aligned} |f(x)-f(c)| &= \left|f(x)-\sum_{n=1}^{\infty}f_N(x)+\sum_{n=1}^{\infty}f_N(x)-\sum_{n=1}^{\infty}f_N(c)+\sum_{n=1}^{\infty}f_N(c)-f(c) \right| \\ &\leq \left|f(x)-\sum_{n=1}^{\infty}f_N(x)\right|+\left|\sum_{n=1}^{\infty}f_N(x)-\sum_{n=1}^{\infty}f_N(c)\right|+\left|\sum_{n=1}^{\infty}f_N(c)-f(c) \right| \\ &= \left|f(x)-\sum_{n=1}^{\infty}f_N(x)\right|+\sum_{n=1}^{\infty}\left|f_N(x)-f_N(c)\right|+\left|\sum_{n=1}^{\infty}f_N(c)-f(c) \right| \\&\leq \frac{\epsilon}{3}+n.\frac{\epsilon}{3n}+\frac{\epsilon}{3} \\&=\epsilon \end{aligned}$$
Thus, $f$ is continuous on $A$.