So I want to prove that $$\lim_{x \to a} \sum_{n=0}^{\infty} f_n(x) = \sum_{n=0}^{\infty} \lim_{x \to a} f_n(x)$$ is true if $\displaystyle \sum_{n=0}^\infty f_n(x)$ converges uniformly.
I'd like to avoid advanced theorems, like Dominated Convergence etc which is all I find on similar questions. Can this be proven in an elementary way?
Let $g_n = \sum_{k=0}^n f_k$, so you want to show
$$\lim_{x\to a} g(x) = \lim_{n\to \infty} \lim_{x\to a} g_n(x).$$
If $g_n$ converges to $g$ uniformly.
Let $$C = \lim_{x\to a} g(x)\ \ \text{ and }\ \ a_n = \lim_{x\to a} g_n(x).$$ You want to show $\lim_{n\to \infty} a_n = C$. Note that for all $x$, we have
$$|a_n -C | \le |a_n - g_n(x)| + | g_n(x) -g(x)| + |g(x) - C|.$$
As $g_n \to g$ uniformly, there is $N\in \mathbb N$ so that $|g_n(x) - g(x)| < \epsilon/3$ for all $n\ge N$ and for all $x$.
By definition of $C$, there is $\delta>0$ so that $|g(x) - C|<\epsilon/3$ for all $x$ such that $|x- a| < \delta$.
Now we deal with the first term. By definition of $a_n$, there is $\delta_n >0$ so that $|g_n(x) - a_n| < \epsilon/3$ for all $x$ so that $0<|x-a|<\delta_n$.
Now if $n\ge N$, let $x_n$ be found so that $0<|x_n-a| < \min\{\delta, \delta_n\}$. Then we have
$$|a_n - C| < \epsilon.$$
This is true for all $n \ge N$. As $\epsilon >0$ is arbitrary, we have
$$\lim_{n\to \infty} a_n = C.$$