$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}} \newcommand{\tV}{\tilde{V}} \newcommand{\tW}{\tilde{W}}$
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 $\mu_4(Y):=\E(Y-\E Y)^4\ll1/(\al+\be)^2$. So, if $X_1,\dots,X_n$ 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) \le \frac{1}{x^4}\sum_1^n \mu_4(X^{(i)}) \ll \frac{1}{x^4}\sum_1^n \frac1{n^2} \to0, \tag{1} \end{equation} as desired.
More generally, suppose now that $Y_1,\dots,Y_n$ are iid r.v.'s with values in a finite interval and a pdf bounded away from $0$. Then the function $\psi:=G^{-1}$ inverse to the cdf $G$ of $Y_1$ is Lipschitz. Also, we can write $Y_i=\psi(X_i)$ (see e.g. page 4) and $Y^{(i)}=\psi(X^{(i)})$, with $X_1,\dots,X_n$ as before. So, for $V:=X^{(i)}$, an independent copy $\tV$ of $V$, $W:=\psi(V)=Y^{(i)}$, and $\tW:=\psi(\tV)$, and some real constant $L>0$, we have the Lipschitz condition $|W-\tW|\le L|V-\tV|$ and hence
\begin{multline*}
\mu_4(Y^{(i)})=\E(W-\E\tW)^4=\E\E(W-\tW|W)^4
\le\E\E\big((W-\tW)^4|W\big) \\
=\E(W-\tW)^4\le L^4\E(V-\tV)^4\le8 L^4\mu_4(V)=16 L^4\mu_4(X^{(i)});
\end{multline*}
here we used (a conditional version of) Jensen's inequality and the inequality $(a-b)^4\le8(a^4+b^4)$ with $a=V-\E V$ and $b=\tV-\E\tV=\tV-\E V$. So, for any real $x>0$, similarly to (1) we have
\begin{equation*}
\P(\sup_{1 \le i \le n} |Y^{(i)} - \E Y^{(i)}| \ge x) \le \frac{1}{x^4}\sum_1^n \mu_4(Y^{(i)})
\le16 L^4\frac{1}{x^4}\sum_1^n \mu_4(X^{(i)}) \ll \frac{1}{x^4}\sum_1^n \frac1{n^2} \to0,
\end{equation*}
as desired.
The condition that the r.v.'s $Y_i$ take values in a finite interval and have a pdf bounded away from $0$ cannot be dropped. E.g., suppose that $\P(Y_i=0)= \P(Y_i=1)= 1/2$. Then for $m:=\lfloor n/2\rfloor$ we have $p_m:=\P(Y^{(m)}=0)=\P(B_{n,1/2}\ge m)\to1/2$ as $n\to\infty$, where $B_{n,1/2}$ is a binomial r.v. with parameters $n,1/2$. So, $\E Y^{(m)}=\P(Y^{(m)}=1)=1-p_m\to1/2$ and hence $|Y^{(m)}-\E Y^{(m)}|\to1/2\ne0$; the convergence here is in probability (and even almost surely).
Note also that this two-point distribution of $Y_i$ can be appropriately approximated, however closely, by an absolutely continuous distribution with pdf taking zero (or arbitrarily close to zero) values on a subinterval of the interval $[0,1]$ of length arbitrarily close to $1$.