Pick an origin for $\Bbb E^n$, and call the resulting vector space $\Bbb R^n$. To save notation, let's put the origin in $S$, so that $S$ now becomes a (vector) subspace. Write $\Bbb R^n=S\oplus S^\perp$. Now, identifying points $x\in\Bbb E^n$ with the corresponding vectors in $\Bbb R^n$, write $x=x_1+x_2$, where $x_1\in S$ and $x_2\in S^\perp$. Then $\pi_S(x)=x_1$ and $x-\pi_S(x) = x_2\in S^\perp$.
Here's a lemma you need to prove, using dot products. (It's just a higher-dimensional Pythagorean Theorem.)
Lemma: Writing $x=x_1+x_2$, $x_1\in S$, $x_2\in S^\perp$, we have $\|x\|^2 = \|x_1\|^2 + \|x_2\|^2$.
The vector $\overrightarrow{x\pi_S(x)}$ can now be written as $\pi_S(x)-x=-x_2$, so our formula for $R_S$ becomes $R_S(x) = (x_1+x_2)+2(-x_2)= x_1-x_2$. Now it's clear that for any $x,y\in\Bbb R^n$, we have
$$R_S(x)-R_S(y) = (x_1-x_2)-(y_1-y_2) = (x_1-y_1) + (y_2-x_2).$$
Because $x_1-y_1\in S$ and $y_2-x_2\in S^\perp$, we have
\begin{align*}
\|R_S(x)-R_S(y)\|^2 &= \|x_1-y_1\|^2 + \|y_2-x_2\|^2 = \|x_1-y_1\|^2 + \|x_2-y_2\|^2 \\&= \|(x_1+x_2)-(y_1+y_2)\|^2 = \|x-y\|^2,
\end{align*}
as desired.