Let $F$ be, say, a non-archimedean local field. Let $G$ be a connected reductive (can be assumed simply connected) quasi-split group $G$ over $F$. Let 
$X\in\operatorname{Lie}G$ be semisimple and $G\_X:=C\_G(X)$. How can we give an explicit example in which $G\_X$ is not quasi-split? 


Thanks and pardon for the probably pretty simple question. I don't really know if such $G\_X$ exist, but will be surprised if they don't, as I learned that when Langlands and Shelstad did descent for transfer factors they had to deal with the possibility of non-quasi-split centralizer (though for the group).



Twisted Levi of a quasi-split group that is not quasi-split
The answer is no. Take $F$ to be reals, and consider the subgroup $K=U(g)\subset Sp\_{2g}(F)$. Then $K$ is the centraliser of "multiplication by $i$", and is not quasi-split since it is compact. The element multiplication by $i$ on $\mathbb{C}^g$ is to be viewed as the $g$-fold direct sum of the two by two matrix $\begin{pmatrix} 0 & 1 \cr -1 & 0\end{pmatrix}$ on the real vector space $(\mathbb{R}^2)^g$. 


A similar example can be given over non-archimedean local fields. The resulting group will not be anisotropic over $F$, but will not be quasi-split. To see this, let $E/F$ be a quadratic extension and denote by $x\mapsto \overline{x}$ the action by the non-trivial element of the Galois group. Let $h$ be a non-degenerate Hermitian form with respect to $E/F$ on the $E$ vector space $E^g$. The relative norm one elements $S$ of $E/F$ lie in $U(h)$. The "imaginary part" $\Omega$ of $h$ is a non-degenerate symplectic form on $F^{2g}=E^g$ and the centraliser of a non-zero element $X\in Lie (S)\subset Lie Sp\_{2g}$ is precisely $U(h)$. For a suitable choice of $h$, the group $U(h)$ is not quasi split. (If the number of variables $g\geq 3$, the Hermitian form does represent a zero for non-archimedean local fields, so the group $U(h)$ is isotropic).


