According to (a special case of) Chevalley's theorem: Let $G$ be a closed algebraic subgroup of $GL_n(\mathbb C)$ and $H$ a closed subgroup of $G$, then there exists a line (through the origin) $L\subset \mathbb C^n$ such that $H=\{g\in G; \ g\cdot L=L\}$.
Now consider a compact variety $X$ such that $G/H\hookrightarrow X$ be equivariantly embedded. hence we have the situation $G/H=G\cdot x\overset{ open}{\hookrightarrow}\overline{G\cdot x}\overset{ closed }{\hookrightarrow}X$. So if $Gx$ is a mininmal dimension orbit then it is compact? Is my assertion correct?
You are assuming beforehand that $X$ is a compact variety on which $G$ acts, in which $G/H$ embeds equivariantly onto an orbit of minimal dimension in $X$? Then yes, your claim follows because orbits of minimal dimension are closed, and a closed subset of a compact variety is compact.
By the way, I don't think you have stated Chevalley's theorem correctly. Rather, given $H$ as a closed subgroup of a linear algebraic group $G$, there exists an embedding of $G$ into some $\operatorname{GL}_n(\mathbb C)$ and a line $L \subset \mathbb C^n$ such that $H = \{g \in G: g.L = L\}$