= Solution
Choose a linear functional $\ell:E\to\mathbb R$ which is nonzero on every <root>; one exists because there are only finitely many <root> hyperplanes to avoid. Declare a <root> positive when $\ell(\alpha)>0$, and negative when $\ell(\alpha)<0$. This orders the <roots> by sign, compatibly with addition whenever the sum is a <root>. One can refine it to a lexicographic total order by completing $\ell$ to a coordinate system. Since negation permutes the <roots>,
$$
\boxed{R=R^+\sqcup R^-,\qquad R^-=-R^+.}
$$
A <simple root> is a <positive root> not expressible as the sum of two <positive roots>. If distinct <simple roots> $\alpha,\beta$ had $\alpha-\beta$ as a <root>, it would be either positive or negative. In the first case $\alpha=\beta+(\alpha-\beta)$ decomposes $\alpha$ into two <positive roots>; in the second $\beta=\alpha+(\beta-\alpha)$ decomposes $\beta$. Both contradict simplicity. If they are equal, their difference is zero and is also not a <root>. Hence \b[the difference of two <simple roots> is never a <root>].
For use in the final subpart, the <simple roots> form a <basis>, and every <positive root> is a nonnegative integer combination of them. Here are the needed reasons. Repeatedly decompose a nonsimple <positive root> into two <positive roots>; their $\ell$ values decrease, and the finite <root> set makes this terminate. This gives the integer combinations. Distinct <simple roots> have nonpositive <inner product>: a positive <inner product> would make $\alpha$ and $-\beta$ obtuse, so the qualified root-sum lemma would make $\alpha-\beta$ a <root>, just ruled out. Finally, a nontrivial linear relation between <simple roots> can be split into positive and negative coefficients, giving the same vector $u$ as positive combinations of disjoint sets. Its squared norm, computed using those two expressions, is nonpositive, whereas a nonzero positive combination has positive $\ell$ value. This contradiction proves linear independence. Their span is $E$ because they generate all <roots>.
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