# Uniform equicontinuity from pointwise equicontinuity over compact set

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}