Let G be a connected reductive group over complex numbers whose derived subgroup is simply connected. Let u be a unipotent element of G. The centralizer of u in G is denoted by Z_(u). Let F_(u) be a maximal connected reductive subgroup of Z_(u). My question is whether the derived subgroup of F_(u) is simply connected.
Looking at examples which involve classical groups, I'd say that the answer to your question is no. A useful source to consult is the article on "Conjugacy classes" by Springer and Steinberg in the 1970 Lecture Notes in Mathematics No. 131 (IAS seminar writeups), especially part IV.2 of the article. Here they describe for classical groups explicit Levi factors in centralizers of unipotent or nilpotent elements (which are mostly interchangeable in characteristic 0). For example, if you start with a symplectic group (simply connected), you typically get a Levi factor which is a product of various symplectic and special orthogonal groups. The latter are not simply connected, however.
(In most of this discussion, the characteristic of the field isn't important.)
P.S. It's usually a delicate matter in a given simply connected semisimple group $G$ to decide when the derived subgroup of some Levi subgroup in a proper connected subgroup of $G$ is itself simply connected. (Even leaving aside in prime characteristic the more delicate question of whether such Levi subgroups exist.) On the positive side, parabolic subgroups are wellbehaved: they always have Levi decompositions (with all Levi subgroups conjugate in $G$) and the derived subgroup of such a Levi is always simply connected. But the study of various centralizers in $G$ is usually not at all straightforward.
I'm emphasizing arbitrary characteristic to underline the fact that for algebraic groups the concept simply connected is usually understood in terms of the position of a root lattice in a weight lattice rather than just in terms of covering groups. For example, in their paper Groupes reductifs and the later complements, BorelTits constructed a split group in a certain way inside a nonsplit group; but even when the original group is simply connected it requires a nonobvious argument with roots and weights to see that their split subgroup is simply connected.
The short answer to this question is NO (to give a more precise answer one would have to go into much detail and invoke the classification of nilpotent orbits in simple Lie algebras). Probably, the quickest way to see this is to look at the last section of the SpringerSteinberg paper on conjugacy classes (published in Springer LNM, vol. 131). There one can find a description of the reductive parts of the centralisers $C_G(u)$ in the case where $G$ is a classical group. If we take for $G$ a symplectic group of rank $n$ (which is simply connected), then the conjugacy classes of unipotent elements in $G$ are parametrised by certain partitions of $2n$. The reductive part $L$ of $C_G(u)$ decomposes as a direct product of orthogonal ans symplectic groups. The derived subgroup of $L$ will not be simply connected if one of the orthogonal groups $O_m$ with $m\ge 5$ occurs as a direct factor of $L$. This condition can be described in purely combinatorial terms and if $n$ is sufficiently large there will be many such instances.

$\begingroup$ I've just reread Jim's answer and realised that he has already suggested this example. Sorry for doubleposting! $\endgroup$ – Alexander Premet Sep 11 '12 at 10:34
You can also see the book by Jantzen to find what Professor Premet said. The book is Nilpotent orbits in representation theory pages 1221, in ''Lie Theory, Lie algebras and representations'', Progress in Mathematics 228, Birhauser, Boston.Basel.Berlin, 2004.
You can find it in Proposition 2,page26 of above book. Maybe that will give you a precise explanation about the structure of Centralizer of nilpotent element.