I came up with a simpler proof of theorem 7.11 from Rudin PMA but I feel unsure that my reasoning is sound. Am I allowed to apply the limit inequality theorem as I have on the line marked (1)?
My concern is that I do not know if the limits exist which is necessary to use this theorem.
Theorem 7.11: Suppose $f_n \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$\lim_{t\to x} f_n(t) = A_n \qquad (n \in N).$$ Then $\{A_n\}$ converges, and $$\lim_{t\to x} f(t) = \lim_{n\to \infty} A_n.$$ In other words, the conclusion is that
$$\lim_{n\to\infty}\lim_{t\to x}f_n(t)=\lim_{t\to x}\lim_{n\to \infty}f_n(t)$$ My proof:
Fix $\epsilon > 0$. Then by uniform convergence, there exists some $N\in \mathbb{N}$ such that $\forall n>N$ $$|f_n(t)-\lim_{n\to \infty}f_n(t)|<\epsilon \qquad \forall t\in E$$ Now we may write this inequality as
$$\lim_{n\to \infty}f_n(t)-\epsilon< f_n(t)<\lim_{n\to \infty}f_n(t)+\epsilon$$
Now since this holds $\forall t \in E$, we have by the limit inequality theorem that
$$\lim_{t\to x}(\lim_{n\to \infty}f_n(t)-\epsilon)< \lim_{t\to x}f_n(t)<\lim_{t\to x}(\lim_{n\to \infty}f_n(t)+\epsilon) \qquad (1)$$
which implies that
$$|\lim_{t\to x}f_n(t)-\lim_{t\to x}\lim_{n\to \infty}f_n(t)|<\epsilon$$
Therefore, we have shown that $$\lim_{n\to\infty}\lim_{t\to x}f_n(t)=\lim_{t\to x}\lim_{n\to \infty}f_n(t)$$