Connectedness of stabilizer of regular element Let $\mathfrak{g}$ be a complex simple Lie algebra and $x \in \mathfrak{g}$ be a regular element, i.e. its centralizer is of minimal dimension.
Consider the adjoint action of the adjoint group $G$ (with trivial center) on its Lie algebra $\mathfrak{g}$.
Is it true that the stabilizer of $x$ in $G$ is always connected?
If $x$ is semisimple, then it is known that the stabilizer of $x$ in the simply connected group is connected (see the book of Collingwood-McGovern, Theorem 2.3.3), and so by projection, also in the adjoint group.
If $x$ is regular and nilpotent (also called principal nilpotent), then one can check in the lists of all Lie algebras in the book of Collingwood-McGovern to check that the stabilizer is connected.
But what about a general regular element?
And if it is true, is there a simple proof (without using the classification of simple Lie algebras)?
 A: $\def\semi{_{\text{semi}}}\def\nil{_{\text{nil}}}$This is just an elaboration of my comment.  If $X \in \mathfrak g$ is regular, with semisimple part $X\semi$ and nilpotent part $X\nil$, then $X\nil$ is regular nilpotent in $\operatorname{Lie}(\operatorname C_G(X\semi))$, so you have said that you already know that $\operatorname C_{\operatorname C_G(X\semi)^\circ}(X\nil)$ is connected.  On the other hand, $\operatorname C_G(X\semi)$ is connected.  (Your argument for this by reduction to the simply connected case doesn't work, but it is true in general, in characteristic 0 or even just in not-too-small positive characteristic.  The reference that I know is Section 7 of Yu - Construction of tame supercuspidal representations, although that's clearly the wrong place to look for general questions of this sort; better to look in Steinberg.  (Probably there's a precise reference in Collingwood–McGovern; the "simple connectedness implies connectedness" result you cite, which is valid on the group as well as the Lie-algebra level, is due to Steinberg.)  Thus, $\operatorname C_{\operatorname C_G(X\semi)^\circ}(X\nil)$, which we have seen is connected, equals $\operatorname C_{\operatorname C_G(X\semi)}(X\nil) = \operatorname C_G(X)$.
