 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,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 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.