Fix $\epsilon>0$ and consider the interval $[0,T]$, for $T>0$ and a discretization of it: fix $\delta=\delta(\epsilon)>0$, the discretization parameter and consider the subintervals $[K\delta, (K+1)\delta]$ for $K=0...T/\delta-1$ (without being precise on the truncations for simplicity).
Suppose to have $u^\epsilon, u \colon [0,T] \to \mathbb{R}$ continuous functions, with $u^\epsilon$ depending on $\epsilon$ in some way.
Suppose that I have shown for every $K=0...T/\delta-1$: $$sup_{t \in [K\delta, (K+1)\delta]} |u^\epsilon(t)-u(t)| \leq C \epsilon$$ for some constant $C$ independent of $\epsilon$ (and so independent also of $\delta$) and also $C$ independent of $K$
Does it follow immediately that: $$sup_{t \in [0,T]} |u^\epsilon(t)-u(t)| \leq C \epsilon$$ right?
Yes it does. Let $t \in [0, T]$ be arbitrary. There exists $K$ such that $t \in [K\delta, (K + 1)\delta]$. Thus $|u^{\varepsilon}(t) - u(t)| \leq C\varepsilon$. Taking the sup over $t$ yields $\sup_{t \in [0, T]}|u^{\varepsilon}(t) - u(t)| \leq C\varepsilon$.
More generally, if $\{A_\lambda : \lambda \in \Lambda\}$ is a collection of subsets of $[-\infty, \infty]$, then $$\sup(\bigcup_{\lambda \in \Lambda}A_\lambda) = \sup_{\lambda \in \Lambda}\sup(A_\lambda).$$