$\def\semi{_{\text{semi}}}\def\nil{_{\text{nil}}}$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 $\operatorname C_{\operatorname C_G(X\semi)^\circ}(X\nil) = \operatorname C_G(X)$.

$\def\semi{_{\text{semi}}}\def\nil{_{\text{nil}}}$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 $\operatorname C_{\operatorname C_G(X\semi)}(X\nil) = \operatorname C_G(X)$.