Let $x:=(x_1,\dots,x_n)$ and $y:=(y_1,\dots,y_n)$. By compactness and continuity, the maximum of
$P:=P(x,y):=\prod_{i=1}^{n}|x_i-y_i|$ under the given conditions on $x,y$ is attained. In the sequel, let $(x,y)$ be a point of attainment of this maximum. Then clearly $x_i\ne y_i$ for all $i$.
Some terminology: Let us say that a subset $J$ of the set $[n]:=\{1,\dots,n\}$ is connected if it is the intersection of $[n]$ with an interval. An $x$-run is a maximal connected set of constancy of the function $x\colon i\mapsto x_i$. An $(y<x)$-run is a maximal connected subset $J$ of $[n]$ such that $y_i<x_i$ for all $i\in J$. Similarly defined are the $y$-runs and $(y>x)$-runs.
Without loss of generality (wlog) $x_n<y_n$.
Take any $(y<x)$-run $J$; then $J\subseteq[n-1]$. If $\{i-1,i\}\subseteq J$ and $y_{i-1}<y_i$, then, replacing $y_i$ and $y_n$ respectively by $y_i-h$ and $y_n+h$ with any $h\in(0,y_i-y_{i-1}]$ results in a greater value of $P$, which contradicts the maximality of $(x,y)$. So, $y_i$ is constant on $J$. So, by the AM-GM inequality, $x_i$ is constant on $J$ as well.
Take now the rightmost $(y<x)$-run $J_*$, with $j_*:=\min J_*$ and $k_*:=\max J_*$. Then the minimal value, say $\mu$, of $y-x$ to the right of $J_*$ (that is, on the set $\{k_*+1,\dots,n\}$) is strictly positive. Take any $(x<y)$-run $K\subseteq[j_*-1]$. If $\{i-1,i\}\subseteq K$ and $x_{i-1}<x_i$, then, replacing $x_i$ by $x_i-h$ and $x_j$ for each $j\in \{j_*,\dots,n\}$ by $x_i+h/(n-j_*+1)$ with any $h\in(0,(x_i-x_{i-1})\wedge\mu]$ results in a greater value of $P$, which contradicts the maximality of $(x,y)$. So, $x_i$ is constant on $K$. So, by the AM-GM inequality, $y_i$ is constant on $K$ as well.
Thus, on each $(y<x)$-run and each $(y>x)$-run, except maybe on the rightmost $(y>x)$-run $K_*$, $x$ and $y$ are each constant. Moreover, it is easy to see that the rightmost $y$-run $Y_*$ is the union of the rightmost $(y<x)$-run $J_*$ and rightmost $(y>x)$-run to the left of $K_*$. So, if the constant value $y$ on $Y_*$ is $>0$, then, replacing $y$ by $y-|K_*|h$ on $Y_*$ and $y$ by $y+|Y_*|h$ on $K_*$ for small enough $h>0$ results in a greater value of $P$, which contradicts the maximality of $(x,y)$. So, $y=0$ on $Y_*$ and hence to the left of $Y_*$, so that $y_i=0$ for $i=1,\dots,k$, where $k:=\max Y_*$.
Thus, $x_i=u$ and $y_i=0$ for $i=1,\dots,k$, and $y_i>x_i$ for $i=k+1,\dots,n$, with $\sum_{k+1}^n(y_i-x_i)=\sum_{k+1}^n y_i-\sum_{k+1}^n x_i=\sum_1^n y_i-\sum_{k+1}^n x_i=n-(n-ku)=ku$. So, by the AM-GM inequality, $\prod_{k+1}^n|y_i-x_i|=\prod_{k+1}^n(y_i-x_i)\le(\frac{ku}{n-k})^{n-k}$. Moreover, $n=\sum_1^n x_i\ge nu$, so that $0\le u\le1$ and \begin{equation} \prod_1^n|y_i-x_i|=\prod_1^k|y_i-x_i|\prod_{k+1}^n|y_i-x_i|\le u^k\,(\frac{ku}{n-k})^{n-k} \le(\frac{k}{n-k})^{n-k}=f(t)^n, \end{equation} where \begin{equation} f(t):=(\frac{t}{1-t})^{1-t}\le e^c \end{equation} for $t\in(0,1)$, where $c=0.278\dots<1/2$, as desired.