I have a two part question:
(1). Let $H$ be a maximal toral subalgebra of a semisimple lie algebra $L$ and $\Phi$ be the associated root system. Prove that there exists $h\in H$ such that $\alpha(h)\neq 0$ for all $\alpha\in \Phi$.
(2). Deduce using the above fact that every maximal toral subalgebra is the centralizer of some $1$-dimensional subalgebra of $L$.
Please help.
I have been thinking about this question for a little while, and I think I just hit upon the answer.
For part (1), consider that, by the non-degeneracy of the Killing form on a semisimple Lie algebra, $\ker \alpha \neq H$ (where we identify $\Phi \subset H^*$ as you mentioned in the comments, so $\alpha \colon H \to \mathbb{C}$). Then, $\dim \ker \alpha < \dim H$ for all $\alpha \in \Phi$. It is a common fact from linear algebra that a vector space cannot be written as a union of finitely many proper subspaces, so take $$h \in H \setminus \left(\bigcup_{\alpha \in \Phi} \ker \alpha\right) \neq \emptyset$$ Then, by construction, it must be that $\alpha(h) \neq 0$ for all $\alpha \in \Phi$.
Now that we have part (1), the second part follows. As part of the proof that the Killing form is non-degenerate on $H$, we used that $C_L(H) = H$. Take $h$ as above. Consider $C_L(h)$, that is, all elements $x \in L$ such that $[x,h]=0$. We immediately get $H = C_L(H) \subset C_L(h)$. Now, take an arbitrary element $x \in C_L(h)$, which can be written as $x = h' + \sum_{\alpha \in \Phi} x_\alpha$ for $h' \in H$ and $x_\alpha \in L_\alpha$ by the root space decomposition of $L$. Then, $$0 = [h'+\sum_{\alpha \in \Phi} x_\alpha, h] = [h',h]+\sum_{\alpha \in \Phi}[x_\alpha,h] = \sum_{\alpha \in \Phi} \alpha(h) x_\alpha$$ From this, it must be that $x = h' \in H$, so $C_L(h) \subset H$ and thus $C_L(h) = H$.