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 $x=\left( \begin{array}{cc} 1 & 1 \\\ 0 & 1 \\\ \end{array} \right)$ is $X$. (Hence it is closed.) Let $H$ be the subgroup of matrices of the form $\left( \begin{array}{cc} a & 0 \\\ 0 & a^{-1} \\\ \end{array} \right) \times \left( \begin{array}{cc} a & 0 \\\ 0 & a^{-1} \\\ \end{array} \right)\subset G$. Then $Hx=\left(\begin{array}{cc} 1 & b \\\ 0 & 1 \\\ \end{array} \right),$ for $b\ne 0.$ Hence $Hx$ is not closed.

$\def\smat#1{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}$

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.