$\newcommand{\ep}{\varepsilon}$
Let $d_n:=\|X-Y\|_\infty$, where $X=(X_1,\dots,X_n)$ and $Y=(Y_1,\dots,Y_n)$ are independent random points each uniformly distributed in $[0,1]^n$, so that $X_1,\dots,X_n,Y_1,\dots,Y_n$ are independent random variables (r.v.'s), each uniformly distributed in $[0,1]$. Then for any fixed $\ep\in(0,1)$, using the condition that $X_1,\dots,X_n,Y_1,\dots,Y_n$ are independent and identically distributed, we have

\begin{equation}
\begin{aligned}
P(d_n<1-\ep)&=P(\max_{i\le n}|X_i-Y_i|<1-\ep) \\
&=P(|X_1-Y_1|<1-\ep,\dots,|X_n-Y_n|<1-\ep) \\
&=P(|X_1-Y_1|<1-\ep)\cdots P|X_n-Y_n|<1-\ep) \\
&=P(|X_1-Y_1|<1-\ep)^n\to0
\end{aligned} \tag{1}
\end{equation}
(as $n\to\infty$), since $P(|X_1-Y_1|<1-\ep)<1$. That is, $d_n\to1$ in probability.

More specifically, we have $P(|X_1-Y_1|<1-\ep)=1-\ep^2$.
So, taking any real $c>0$ and then letting $\ep=\sqrt{c/ n}$, we see that for $n>c$ formula (1) implies
\begin{equation*}
P(n(1-d_n)^2>c)
=P(d_n<1-\sqrt{c/ n})=(1-c/n)^n\to e^{-c}=P(Z>c),
\end{equation*}
where $Z$ is a r.v. with the standard exponential distribution; that is,
$n(1-d_n)^2$ converges in distribution to $Z$. Informally, this can be written as
$n(1-d_n)^2\approx Z$ and hence
\begin{equation*}
d_n\approx1-\sqrt{Z/n}\approx 1.
\end{equation*}