A group action on an infinite set with the cofinite topology Let $X$ be an infinite set with the co-finite topology and $\varphi:G\times X\to X$ is a continuous group action.
Let $\mathcal{U}=\{U_1, U_2, \ldots, U_n\}$ be a finite open cover of $X$. It is clear that $A=\cup_{i=1}^n X-U_i$ is a finite set. 
Are there $x\neq y\in U_i$, for some $i\in \{1, 2, \ldots, n\}$, such that $(Gx\cup Gy)\cap A=\emptyset$?
 A: Well, for all $i \in \{1,...,n\}, (X - U_i)$ is a finite set (by the definition of cofinite topology) so $A$ is just a finite union of finite sets : it must be finite as well.
For the second question, $(Gx \cup Gy) \cap A = \emptyset$ is equivalent to : there exist $x \not = y \in U_i$ such that for all $g \in G$, $gx$ and $gy$ are in all the $U_i$.
So, it such $x$ and $y$ were to exist, you would need to have $x$ and $y$ be members of all the $U_i$, and that the intersection of all $U_i$ is stable under the action of $G$. Sometimes it will be possible, sometimes not : I will give you two examples.
In the first example, just consider the trivial action on $X = \mathbb{N}$ with $G = \{1\}$. Then $\mathcal{U} =\{\mathbb{N} - \{1\},\mathbb{N} - \{0\}\}$ is a finite open cover of $X$,$A = \{0,1\}$ and you can easily check that $x = 2$,$y = 3$ verifies the condition.
In the second example, we again take $X = \mathbb{N}$, but this time we take $G$ to be the set of homeomorphisms of $X$ to itself : it can easily be checked that $G$ is the set of all bijections from $X$ to itself (indeed, a bijection sends finite sets to finite sets, and so does its inverse). Then, for all $x \in X$, $Gx = X$ (I will let you check that you can find a bijection sending $x$ to any element of $\mathbb{N}$). And so, whatever the choice of the finite open cover is, you will have $(Gx \cup Gy) \cap A = A$, and $A$ is not equal to the empty set as long as there exists a $U_i$ different from $X$.
So in order to check your result, you need more details on $X$ and the finite cover !
I may have made a mistake though, I worked this out on the fly ; let me know about any problem/question.
