Suppose that a sequence of equi-continuous functions $\{f_n\}_{n \in \mathbb{N}}$ is such that for the fixed constant $K > 0$, each $f_n: [0, K] \to [0, K]$ and it is strictly increasing.
Take any $x \in (0,K)$. Is it possible to prove that there exists some $\epsilon = \epsilon(x) > 0$ such that by making $\delta = \delta(\epsilon) > 0$ sufficiently small, we have \begin{equation} f_n(x) - \epsilon \geq f_n(x-\delta), \end{equation} for all $n$?
No, take $f_n(x)=x/n$ in $[0,K]$ for $n\geq 1$. The sequence is equi-continuous and the functions are strictly increasing. On the other hand, for $\epsilon>0$, $$f_n(x) - \epsilon \geq f_n(x-\delta)$$ if and only if $n\epsilon \leq \delta$ for any $n\geq 1$. This is a contradiction because the left-hand side is unbounded.