Need help understanding proof that there always exists a closed orbit. I know every orbit $G.x$ is open in $\overline{G.x}$, and if we assume that $G.x$ is not closed then $\overline{G.x}\setminus G.x$ is a union of orbits.
It is then claimed that the dimension of $\overline{G.x}\setminus G.x$ is strictly less that the dimension of $G.x$. I want to know know how is this true, and also how does it follow from this that the orbit with minimal dimension has to be closed?
 A: Proposition:Let $X$ be a variety, $U$ a dense open subset, and $Z$ the complement of $U$ in $X$.  Then $Z$ has strictly lower dimension than $X$.  
Proof: Let $X_1, ... , X_t$ be the irreducible components of $X$, and let $Z_1, ... , Z_s$ be the irreducible components of $Z$.  So the dimension of $X$ (resp. $Z$) is the maximum of the dimension of the $X_i$ (resp. $Z_i$).  
Since $U$ is dense in $X$, $U$ has nonempty intersection with each $X_i$.  For if, say, $U \cap X_1 = \emptyset$, then $U$ and hence $\overline{U} = X$ is contained in the closed set $X_2 \cup \cdots \cup X_t$, which is properly contained in $X$.
Since $Z_i$ is irreducible, it is contained in one of the maximal irreducible subsets of $X$, say $X_j$.  And the containment must be proper, because $U \cap X_j$ is nonempty.  Now a proper, closed, irreducible subset of an irreducible variety must be of strictly lower dimension.  So each of the $Z_1, ... , Z_s$ has strictly less dimension than that of some irreducible component of $X$.  $\blacksquare$
Now take $X = \overline{Gx}, U = Gx$.
N.H.'s answer explains the rest: there exist orbits of minimal dimension, and any such orbit $Gx$ must be closed.  Otherwise, $\overline{Gx} - Gx$ is a union of orbits of strictly lower dimension.
A: By hypothesis $Z_1 := \overline {Gx} \backslash Gx$ is a proper closed subset in $\overline {Gx}$, so $\dim Z_1 < \dim \overline{Gx}$. 
For the induction step : let $x_1 \in Z_1$, if $Gx_1$ is closed the claim is proved, else you can take again $Z_2 = \overline{Gx_1} \backslash Gx_1$, pick a point $x_2 \in Z_2$ and so one. If this procedure doesn't stop for any $Z_i$ of positive dimension, you will end with a $G$-invariant set of dimension zero which certainly contains a closed orbit. 
