In characteristic zero, the connected closed subgroups of $G$ are in 1-1 correspondence with the Lie subalgebras of ${\rm Lie}(G)={\mathfrak g}$, and <strike>the Killing form</strike> a suitably chosen symmetric bilinear $G$-equivariant form on ${\mathfrak g}$ is non-degenerate. <i>In fact we can choose this form to be the trace form for a rational representation for $G$.</i> Since <strike>the Killing form</strike> this form is $G$-equivariant, it is $H$-equivariant, and irreducible summands for non-isomorphic $H$-submodules are orthogonal. Hence the <strike>Killing</strike> form on ${\mathfrak g}$ restricts to a non-degenerate form on ${\mathfrak g}^H={\rm Lie}(Z_G(H)^\circ)$.

I now claim that if $K$ is a connected algebraic group with a non-degenerate trace form $\kappa$ arising from a rational representation $\rho:K\rightarrow {\rm GL}(V)$, then $K$ is reductive. Indeed, if the unipotent radical $R_u(K)$ is non-trivial then $\rho(R_u(K))$ is a unipotent subgroup of ${\rm GL}(V)$, so after conjugation is contained in the subgroup of upper-triangular unipotent matrices, so the restriction of $\kappa$ to ${\mathfrak u}={\rm Lie}(R_u(K))$ is zero. Since ${\mathfrak u}$ is an ideal of ${\rm Lie}(K)$, this contradicts the non-degeneracy of the form. In particular, $Z_G(H)^\circ$ is reductive.