$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$
It is easy to see (say, using the formulas for the variance and excess kurtosis for the beta distribution) that, if $Y$ has the beta distribution with parameters $\al,\be>0$, then $\mu_4(Y):=\E(Y-\E Y)^4\ll\al\be/(\al+\be)^5$. So, if $X_1,X_2,\dots$ are iid uniformly distributed on $[0,1]$, then for any real $x>0$ \begin{equation} \P(\sup_{1 \le i \le n} \De^{(i)} \ge x) \le \sum_1^n \P( \De^{(i)} \ge x) \leq \frac{1}{x^4}\sum_1^n \mu_4(X^{(i)}) \ll \frac{1}{x^4}\sum_1^n \frac{n^2}{n^5} \to0, \end{equation} as desired.