Here is an attempted proof. We have that $Z_G(X)$ is the full preimage of $Z_H(X)$ under the covering central extension $\operatorname{Ad}$:
$$
1\longrightarrow C\longrightarrow G\overset{\operatorname{Ad}}\longrightarrow H\longrightarrow 1
$$
where $C$ is the discrete center and $H=\operatorname{Ad}(G)$ the adjoint group of $G$. Now:

> **Theorem** ([Kostant 1963](http://www.ams.org/mathscinet-getitem?mr=158024), [Prop. 14](https://books.google.com/books?id=Ae_Ayk5OitsC&pg=PA353)). *If $X$ is regular, then $Z_H(X)$ is connected abelian.*

To deduce that $Z_G(X)$ is also abelian, pick $g_0$ and $g$ there, projecting to $h_0$ and $h_1$. Since $Z_H(X)$ is connected, it contains a continuous path $h_t$ from $h_0$ to $h_1$. Lift that to a continuous path $g_t$ in $Z_G(X)$ starting at $g_0$. Since $Z_H(X)$ is abelian, the path $c_t=g_tg_0g_t^{-1}g_0^{-1}$ satisfies
$$
\operatorname{Ad}(c_t)=h_th_0h_t^{-1}h_0^{-1}=1.
$$
So $c_t$ lies in the discrete center $C$, hence is constantly equal to $c_0=1$. So $g_0$ commutes with $g_1$. But $\operatorname{Ad}(g)=h_1=\operatorname{Ad}(g_1)$ shows that $g=g_1z$ for some $z\in C$. So $g_0$ commutes with $g$, too.