I was unable to see why $(\alpha+\beta,\alpha) \ne 0$ and $(\alpha+\beta,\beta)\ne0$ implies $\alpha+\beta \not\in\Phi$. Everything else is fine.
$\quad$*Proposition.* Let $L$ be a simple Lie algebra. $H$ and $\Phi$ as above. Then $\Phi$ is an irreducible root system in the sense of $\text{(10.4)}.$
$\quad$Proof. Suppose not. Then $\Phi$ decomposes as $\Phi_1\cup\Phi_2$, where the $\Phi_i$ are orthogonal. If $\alpha\in\Phi_1$, $\beta\in\Phi_2$, then $(\alpha+\beta,\alpha)\neq0$, $(\alpha+\beta,\beta)\neq0$, so $\alpha+\beta$ cannot be a root and $[L_\alpha L_\beta]=0$. This shows that the subalgebra $K$ of $L$ generated by all $L_\alpha(\alpha\in\Phi_1)$ is centralized by all $L_\beta(\beta\in\Phi_2)$; in particular, $K$ is a proper subalgebra of $L$, because $Z(L)=0$. Furthermore, $L$ is normalized by all $L_\alpha(\alpha\in\Phi_1)$, hence by all $L_\alpha(\alpha\in\Phi)$, hence by $L$ (Proposition $8.4$ ($\rm f$)). Therefore $K$ is a proper ideal of $L$, different from $0$, contrary to the simplicity of $L$. $\quad\boxed{\color{white}{\cdot}}$
The assumption is that any root will be in either $\Phi_1$ or $\Phi_2$. But if $\alpha + \beta$ was in $\Phi_1$ it would be orthogonal to $\beta$, and if it was in $\Phi_2$ it would be orthogonal to $\alpha$.