Let $G$ be a reductive affine algebraic $\mathbb{C}$-group (not necessarily connected). Suppose $X$ is an irreducible affine algebraic set over $\mathbb{C}$ where $G$ acts rationally. Suppose that $H$ is a reductive subgroup of $G$ (again not necessarily connected). Let $x\in X$. 


If the orbit $G\cdot x$ is closed in $X$ (in the ball topology), is the sub-orbit $H\cdot x$ also closed? 


NOTE: Originally, I left off the assumption that $H$ is a reductive subgroup, and had not emphasized that I was allowing the adjective "reductive" to include disconnected groups (connected component of identity has trivial unipotent radical). I have editted the problem to reflect my original intentions. 



Closed reductive sub-orbits
$\def\smat#1{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}$


The answer to your question is "no". Let $G=GL(2,\mathbb C) \times GL(2,\mathbb C)$ act on $X=GL(2, \mathbb C)$ by $(A,B)\cdot C= ACB^{-1}.$ The orbit of the matrix $x=\smat{
 1 & 1 \\
 0 & 1 \\\
}$
is $X$. (Hence it is closed.) Let $H$ be the subgroup of matrices of the form
$\smat{
 a & 0 \\
 0 & a^{-1}\\\
}
\times
\smat{
 a & 0 \\\
 0 & a^{-1}\\\
}
\subset G$. Then $Hx=
\smat{
 1 & b \\\
 0 & 1\\\
}$, for $b\ne 0$. Hence $Hx$ is not closed.


