[step:Insert Rademacher signs by pairwise exchangeability]
Enlarge the product space, if necessary, by a further factor carrying independent Rademacher random variables
\begin{align*}
\varepsilon_1,\dots,\varepsilon_n
\end{align*}
that are independent of both samples, and continue to write $\overline{\mathbb E}$ for expectation on this enlarged product space. For each $i\in\{1,\dots,n\}$, the pair $(X_i,X_i')$ has distribution $P\otimes P$, so it has the same distribution as $(X_i',X_i)$. Since the pairs $(X_i,X_i')$ are independent across $i$ and the signs $\varepsilon_1,\dots,\varepsilon_n$ are independent, independently swapping the two coordinates in the $i$th pair according to $\varepsilon_i$ preserves the joint distribution of the full vector of pairs. Therefore the random element
\begin{align*}
\left((X_1,X_1'),\dots,(X_n,X_n')\right)
\end{align*}
has the same distribution as
\begin{align*}
\left((Y_1,Y_1'),\dots,(Y_n,Y_n')\right),
\end{align*}
where for each $i\in\{1,\dots,n\}$,
\begin{align*}
(Y_i,Y_i')
=
(X_i,X_i')
\end{align*}
on the event $\{\varepsilon_i=1\}$, and
\begin{align*}
(Y_i,Y_i')
=
(X_i',X_i)
\end{align*}
on the event $\{\varepsilon_i=-1\}$.
For each fixed finite $\mathcal G\subset\mathcal F$, the measurable map from $(S\times S)^n$ to $[0,\infty]$ given by
\begin{align*}
((x_1,x_1'),\dots,(x_n,x_n'))\mapsto
\sup_{f\in\mathcal G}\left|\frac{1}{n}\sum_{i=1}^{n}\{f(x_i)-f(x_i')\}\right|
\end{align*}
is invariant under replacing the difference in the $i$th coordinate by $\varepsilon_i\{f(X_i)-f(X_i')\}$ after the corresponding coordinate swap. Consequently, for every finite $\mathcal G\subset\mathcal F$,
\begin{align*}
\overline{\mathbb E}\left[\sup_{f\in\mathcal G}|P_nf-P_n'f|\right]
=
\overline{\mathbb E}\left[
\sup_{f\in\mathcal G}
\left|
\frac{1}{n}\sum_{i=1}^{n}\varepsilon_i\{f(X_i)-f(X_i')\}
\right|
\right].
\end{align*}
[/step]