As @JyrkiLahtonen commented, it suffices to show that $t_{\alpha}\in L_i$ and $t_{\beta}\in L_j$, since this implies that $\operatorname{ad}t_{\alpha}\operatorname{ad}t_{\beta}(x)=[t_{\alpha}[t_{\beta}x]]\in L_i\cap L_j=0$ for all $x\in L$, i.e., $\operatorname{ad}t_{\alpha}\operatorname{ad}t_{\beta}=0$, so $$(\alpha, \beta)=\kappa(t_{\alpha}, t_{\beta})=\operatorname{tr}(\operatorname{ad}t_{\alpha}\operatorname{ad}t_{\beta})=\operatorname{tr}(0)=0.$$
To see that $t_{\alpha}\in L_i$, let $\kappa_i$ denote the Killing form of $L_i$. By Humphreys Lemma 5.1, $\kappa_i=\kappa\vert_{L_i\times L_i}$. By Humphreys Corollary 8.2, $\kappa\vert_{H\times H}$ and $\kappa_i\vert_{H_i\times H_i}$ are nondegenerate. This allows us to identify $H$ with $H^*$, by associating to $\gamma\in H^*$ the unique element $t_{\gamma}\in H$ such that $\gamma(h)=\kappa(t_{\gamma}, h)$ for all $h\in H$. Similarly, we can identify $H_i$ with $H_i^*$, by associating to $\delta\in H_i^*$ the unique element $u_{\delta}\in H_i$ such that $\delta(h_i)=\kappa_i(u_{\delta}, h_i)$ for all $h_i\in H_i$. With this in mind, $u_{\alpha}\in H_i$ and $\alpha(h_i)=\kappa_i(u_{\alpha}, h_i)$ for all $h_i\in H_i$. However, by an argument similar to that in the previous paragraph, $\kappa(L_i, L_k)=0$ for $i\neq k$, so given $h=h_1+\cdots+h_t\in H$, $h_k\in H_k$, we have $$\alpha(h)=\alpha(h_i)=\kappa_i(u_{\alpha}, h_i)=\kappa\vert_{L_i\times L_i}(u_{\alpha}, h_i)=\kappa(u_{\alpha}, h_i)=\kappa(u_{\alpha}, h).$$ Then, by uniqueness, $t_{\alpha}=u_{\alpha}\in H_i\subseteq L_i$. Similarly, $t_{\beta}\in L_j$.
Best Answer
A torus $\mathfrak h$ in a simple Lie algebra $L$ is maximal if and only if it is self centralizing: $\mathfrak h = C_L(\mathfrak h)$.
So prove that if a matrix $M \in \mathfrak{sl}_n$ commutes with all diagonal matrices in $\mathfrak{sl}_n$ then $M$ must be diagonal.