The only scalar multiples of a root that are roots are plus or minus the root

Since $H_{c\alpha }=c{H}_{\alpha }$ by the defining equation $B(H_{c\alpha },H)=c\alpha (H)$, the coroots satisfy $h_{c\alpha }=\frac{2c{H}_{\alpha }}{c\alpha (c{H}_{\alpha })}=\frac{2H_{\alpha }}{c\alpha (H_{\alpha })}=h_{\alpha }\mathrm{/}c$.

We first prove that twice a root is never a root. For a root $\gamma$, put $W_{\gamma }=\mathbb{C}{e}_{\gamma }\oplus \mathbb{C}{h}_{\gamma }\oplus {⨁}_{j\ge 1}{\mathfrak{g}}_{-j\gamma }$. This is a finite direct sum because the root spaces are joint eigenspaces for distinct functionals in the finite-dimensional space $\mathfrak{g}$. It is stable under the adjoint action of the triple in [L3]: ${\mathrm{ad}}_{e_{\gamma }}$ and ${\mathrm{ad}}_{f_{\gamma }}$ shift the root-space index by $1$ and $-1$, respectively, the exceptional opposite bracket lands in $\mathbb{C}{h}_{\gamma }$ by [L4], and ${\mathrm{ad}}_{h_{\gamma }}$ preserves every displayed summand.

By [L2] applied to the pair $(\alpha ,\beta )$ we get $2c=\beta (h_{\alpha })=c\alpha (h_{\alpha })\in \mathbb{Z}$, and applied to the pair $(\beta ,\alpha )$ we get $2\mathrm{/}c=\alpha (h_{\beta })=\alpha (h_{\alpha }\mathrm{/}c)\in \mathbb{Z}$.

On $W_{\gamma }$ one has ${\mathrm{ad}}_{h_{\gamma }}=[{\mathrm{ad}}_{e_{\gamma }},{\mathrm{ad}}_{f_{\gamma }}]$, so [L5] makes its trace zero. Its eigenvalues on the displayed direct sum are $2$ on $\mathbb{C}{e}_{\gamma }$, $0$ on $\mathbb{C}{h}_{\gamma }$, and $-2j$ on ${\mathfrak{g}}_{-j\gamma }$. Therefore $0=2-2{\sum }_{j\ge 1}j\dim {\mathfrak{g}}_{-j\gamma }$, so ${\sum }_{j\ge 1}j\dim {\mathfrak{g}}_{-j\gamma }=1$. Hence ${\mathfrak{g}}_{-j\gamma }=0$ for every $j\ge 2$. Applying the same argument to the root $-\gamma$ gives ${\mathfrak{g}}_{j\gamma }=0$ for every $j\ge 2$; in particular $2\gamma$ is not a root.

The two integrality statements say $c=m\mathrm{/}2$ for some integer $m$ and $4\mathrm{/}m\in \mathbb{Z}$, so $m$ divides $4$ and $c\in \{\pm \frac{1}{2},\pm 1,\pm 2\}$.

Now $c\mathrm{\ne }2$ and $c\mathrm{\ne }-2$, since $2\alpha$ and $-2\alpha =2(-\alpha )$ are not roots by step 2.2; and $c\mathrm{\ne }\frac{1}{2},-\frac{1}{2}$, since then $2\beta =\pm \alpha$ would be twice the root $\beta$, again contradicting step 2.2. Hence $c=\pm 1$.