= Solution
Equip $E$ with the <inner product> defining the <root system>, so the <Weyl group> acts by <orthogonal transformations>. As usual, the <root system> spans its ambient space $E$. Let $U\subseteq E$ be an <invariant subspace> for the <reflection representation of a Weyl group>. Its <orthogonal complement> is also invariant.
For a <root> $\alpha$, the <Weyl reflection> satisfies
$$
s_\alpha(u)=u-\frac{2(u,\alpha)}{(\alpha,\alpha)}\alpha.
$$
If $(u,\alpha)=0$ for every $u\in U$, then $\alpha\in U^\perp$. Otherwise choose $u\in U$ with $(u,\alpha)\ne0$. Both $u$ and $s_\alpha(u)$ belong to $U$, and their difference is a nonzero multiple of $\alpha$, so $\alpha\in U$. Hence
$$
R=(R\cap U)\sqcup(R\cap U^\perp).
$$
The two sets are mutually <orthogonal> and individually closed under their root reflections. If both were nonempty, this would decompose $R$ into two <orthogonal> <root systems>, contradicting that it is an <irreducible root system>. All roots therefore lie in one of the two subspaces. Because they span $E$, either $U=E$ or $U^\perp=E$, the latter giving $U=0$. Thus the reflection representation is an <irreducible representation>.
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