Is it true that if all $G^\circ$ - orbits are closed in $X$ then all $G$ - orbits are closed in $X$? Let $G$ be a Lie group acting continuously on a topological space $X$. Let $G^\circ$ be the connected component of the identity element of $G$ and let $[G:G^\circ]$ be finite. Then is the following true? -

If all $G^\circ$ - orbits are closed in $X$ then all $G$ - orbits are closed in $X$

My attempt - 
Let $G\cdot x$ be a $G$ - orbit and let $y\in\overline{G\cdot x}=:Z$. We need to find a $g\in G$ such that $y=g\cdot x$. Now since $X$ is the disjoint union of orbits $y$ is in exactly one of them, say $y\in G\cdot x'$ i.e; $y=g'x'$ for some $g'\in G$. $\implies x'=g'^{-1}y$ and so $x'\in Z$ as $Z$ is $G$ - invariant. I'm not sure how to go from here (or if the statement is false). Also where to use that $[G:G^\circ]$ is finite. 
Thank you.

Let $C_1=G^0,...,C_n$ be the connected component of $G$ and $g_i\in C_i$, $G=\bigcup G^0g_i$. If $x\in X$, the orbit  $G(x)$ is the union $G^0(x), G^0(g_1(x)),....,G^0(g_n(x))$ since $G^0(g_i(x))$ is the orbit of $g_i(x)$ by $G^0$, it is closed, so the orbit of $x$ is closed since it is a finite union of closed subsets.
