Regular elements form a dense Zariski-open subset of a Cartan subalgebra

Each $\alpha$ is a nonzero functional, so each $\ker \alpha$ is a proper subspace, and by [L1] ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ is the complement of finitely many proper subspaces. Induct on the number $k$ of proper subspaces $H_1,\dots ,H_k$. For $k=0$ the assertion is immediate. For $k>0$, choose by induction $u\mathrm{\notin }{\bigcup }_{i<k}{H}_i$ and choose $v\mathrm{\notin }{H}_k$. For each $i<k$ the line $u+tv$ meets $H_i$ for at most one scalar $t$, since two such parameters would imply first $v\in H_i$ and then $u\in H_i$; the same line meets $H_k$ for at most one $t$, since two parameters would imply $v\in H_k$. Because $\mathbb{C}$ is infinite, some $t$ avoids all $k$ exceptional values. Thus a finite union of proper subspaces cannot cover $\mathfrak{h}$, and ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}\mathrm{\ne }\mathrm{\varnothing }$.

Being the complement of a finite union of zero sets of nonzero linear functionals, ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ is Zariski-open. It is the principal open set defined by the nonzero polynomial ${\prod }_{\alpha \in \mathrm{\Phi }}\alpha$ (with empty product $1$); a nonempty principal open subset of an affine space is dense because its coordinate ring is an integral domain.

By [L1] an element $H$ has ${\mathfrak{g}}^H=\mathfrak{h}$ exactly when no root vanishes at $H$, that is, exactly for $H\in {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$; all other elements have strictly larger centralizer dimension. Hence the regular set is the set of elements of $\mathfrak{h}$ with minimal centralizer dimension, and it is nonempty, Zariski-open and dense.