How to show that a sequence of real continuous functions over a compact set that is pointwise equicontinuous and convergent is also uniformly equicontinuous?
I thought to use Arzela-Ascoli theorem to show there is a uniformly convergent subsequence but I can only prove uniform equicontinuity from this if every subsequence is uniformly convergent.
Let $\{f_i: X \to \mathbb{R}\}_{i \in I}$ be a family of real-valued continuous functions on a compact metric space $(X,d)$. Assume that this family is pointwise equicontinuous, i.e. that for every point $x \in X$, we can find for every $\epsilon > 0$ a radius $\delta_x > 0$ such that for every $i \in I$ and every $y \in X$ satisfying $d(x,y) < \delta_x$ we have $\lvert f_i(x) - f_i(y) \rvert < \epsilon$. We prove that this family is uniformly equicontinuous, i.e. that for every $\epsilon > 0$ we can find a $\delta > 0$ such that for every $i \in I$ and $x,y \in X$ with $d(x,y) < \delta$ we have $\lvert f_i(x) - f_i(y) \rvert < \epsilon$.
So fix $\epsilon > 0$ and find for each $x \in X$ a $\delta_x > 0$ such that for every $i \in I$ and every $y \in X$ satisfying $d(x,y) < \delta_x$ we have $\lvert f_i(x) - f_i(y) \rvert < \frac{1}{2}\epsilon$. Define for every $x \in X$ an open subset \begin{aligned} U_x := \{ y \in X \mid d(x,y) < \frac{1}{2} \delta_x\}. \end{aligned} Clearly we have that $x \in U_x$ and thus the open subsets $\{U_x\}_{x \in X}$ cover $X$. Since $X$ is compact, there are finitely many $x_1, \dots, x_n \in X$ such that $U_{x_1}, \dots, U_{x_n}$ cover $X$. Define \begin{aligned} \delta := \frac{1}{2} \min(\delta_{x_1}, \dots, \delta_{x_n}). \end{aligned} We show that for every $i \in I$ and for every $x,y\in X$ with $d(x,y)<\delta$ we have $\lvert f_i(x) - f_i(y) \rvert < \epsilon$. Indeed, we can find some $j \in \{1, \dots, n\}$ such that $x \in U_{x_j}$, meaning that $d(x,x_j) < \frac{1}{2}\delta_{x_j}$. Since we also have $d(x,y) < \delta \leq \frac{1}{2}\delta_{x_j}$, we find \begin{aligned} d(y,x_j) \leq d(y,x) + d(x,x_j) < \frac{1}{2}\delta_{x_j} + \frac{1}{2}\delta_{x_j} = \delta_{x_j}. \end{aligned} Therefore by the choice of $\delta_{x_j}$ we find that for all $i \in I$ \begin{aligned} \lvert f_i(x) - f_i(x_j) \rvert < \frac{1}{2}\epsilon \end{aligned} and \begin{aligned} \lvert f_i(y) - f_i(x_j) \rvert < \frac{1}{2}\epsilon, \end{aligned} so in particular \begin{aligned} \lvert f_i(x) - f_i(y) \rvert \leq \lvert f_i(x) - f_i(x_j) \rvert + \lvert f_i(y) - f_i(x_j) \rvert < \frac{1}{2}\epsilon + \frac{1}{2}\epsilon = \epsilon. \end{aligned}
