I have an idea of the proof of the above statement, but I'm not entirely sure if it's right. Any comments would be appreciated. This question was supposedly answered here but the answer doesn't address the question at all.
Work so far:
Suppose not, i.e. that there exists some $\epsilon > 0$ such that for all $\delta > 0$ and $x,t \in [a,b]$, $|x-t| < \delta$ but $|f_k(x) – f_k(t)| \geq \epsilon$ for some $k \in \mathbb{N}$. Fix this $\epsilon$.
We are given a family of functions $f_n:\mathbb{R} \to \mathbb{R}$ defined on a compact interval $[a,b]$, and that this family is pointwise equicontinuous. Pointwise equicontinuity of {$f_n$} implies that each $f_n$ is continuous, and since the domain is compact, we have that each $f_n$ is uniformly continuous. In particular $f_k$ is uniformly continuous, but this contradicts the hypothesis above.
Best Answer
Let $K \subset \Bbb{R}$ be compact; let $A \neq \varnothing$; let $f_{\alpha}: K \to \Bbb{R}$ for all $\alpha \in A$; let $\{ f_{\alpha} \}_{\alpha \in A}$ be pointwise equicontinuous. For every $c \in K$, if $\varepsilon > 0$ then there is some $\delta_{c} > 0$ such that $x \in B_{K}^{c}(\delta_{c})$ implies $|f_{\alpha}(x) - f_{\alpha}(c)| < \varepsilon/2$ for all $\alpha \in A$. (Here $B_{K}^{c}(\delta_{c})$ denotes the open ball formed over $K$ of center $c$ and radius $\delta_{c}$.) We have $\bigcup_{c \in K}B_{K}^{c}(\delta_{c}/2) = K$, implying by compactness of $K$ that there are some $c_{1},\dots, c_{n} \in K$ such that $$ \bigcup_{j=1}^{n}B_{K}^{c_{j}}(\delta_{c_{j}}/2) = K. $$ If $x \in K$, then there is some $1 \leq j \leq n$ such that $x \in B_{K}^{c_{j}}(\delta_{c_{j}}/2)$ and hence $\in B_{K}^{c_{j}}(\delta_{c_{j}})$; if $|x-y| < \min_{1 \leq i \leq n}\delta_{c_{i}}/2$, then $|y - c_{j}| \leq |y-x| + |x-c_{j}| < \delta_{c_{j}}/2 + \delta_{c_{j}}/2 = \delta_{c_{j}}$, implying that $y \in B_{K}^{c_{j}}(\delta_{c_{j}})$, implying that $$ |f_{\alpha}(x) - f_{\alpha}(y)| \leq |f_{\alpha}(x) - f_{\alpha}(c_{j})| + |f_{\alpha}(c_{j}) - f_{\alpha}(y)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$ for all $\alpha \in A$; this shows the desired uniform equicontinuity.