The Killing length of a root is nonzero

By [L1] choose $e\in {\mathfrak{g}}_{\alpha }$ and $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)\mathrm{\ne }0$ and put $z=[e,f]$. By [L2], $z\in \mathfrak{h}$. For every $H\in \mathfrak{h}$, invariance from [L5] gives $B(z,H)=B(e,[f,H])=\alpha (H)B(e,f)=B(B(e,f)H_{\alpha },H)$. Nondegeneracy of $B{\mathrm{\mid }}_{\mathfrak{h}}$ from [L1] therefore gives $z=B(e,f)H_{\alpha }\mathrm{\ne }0$. Also $[z,e]=\alpha (z)e$ and $[z,f]=-\alpha (z)f$, while $\alpha (z)=B(e,f)\alpha (H_{\alpha })$.

Suppose $\alpha (H_{\alpha })=0$. Then $\alpha (z)=0$, so $[z,e]=[z,f]=0$, the span $\mathfrak{a}=\mathbb{C}e+\mathbb{C}f+\mathbb{C}z$ is a Lie subalgebra with $[\mathfrak{a},\mathfrak{a}]\subseteq \mathbb{C}z$ and $z$ central in $\mathfrak{a}$; in particular $\mathfrak{a}$ is nilpotent and hence solvable by [L4]. Apply the common-eigenvector assertion of [L4] to the adjoint $\mathfrak{a}$-module $\mathfrak{g}$: it gives a one-dimensional invariant subspace $V_1$. Applying it again to the induced action on $\mathfrak{g}\mathrm{/}{V}_1$, and successively to each quotient by the invariant subspaces already obtained, constructs a full invariant flag $0=V_0\subset V_1\subset \cdots \subset V_n=\mathfrak{g}$. In a basis adapted to this flag every ${\mathrm{ad}}_x$, $x\in \mathfrak{a}$, is upper triangular. Hence ${\mathrm{ad}}_z=[{\mathrm{ad}}_e,{\mathrm{ad}}_f]$ is upper triangular with zero diagonal, because the diagonal of a product of upper triangular matrices is the product of their diagonals and scalar diagonal entries commute. Thus ${\mathrm{ad}}_z$ is strictly upper triangular and nilpotent.

But $z\in \mathfrak{h}$, and by [L3] the operator ${\mathrm{ad}}_z$ is semisimple; an operator that is both semisimple and nilpotent is zero, so ${\mathrm{ad}}_z=0$ and $z$ lies in the center. By [L5] the center is zero, so $z=0$, contradicting step 1.1, and therefore $\alpha (H_{\alpha })\mathrm{\ne }0$; because $B(H_{\alpha },H_{\alpha })=\alpha (H_{\alpha })$ by [L1], this is the claim.