Q:

Question about the non-hausdorffness of the cofinite topology

I'm reading a proof about that disproves that the cofinite set 
$$O_c= \{\emptyset,X\}\cup\{S\subset X|\quad X-S\quad \text{is finite }\}$$
is hausdorff. It goes as follows :

Let $U,V \subset X$ such that $U\cap V =\emptyset$, then this implies that $V\subset X-U$ and $X-U$ is finite which implies that $V$ is finite. But then $X-V$ is infinite, contradicting the fact that $V$ is open. 

I don't see how this contradicts the fact that $V$ is open. Is the implication here that infinite sets cannot be open? 

A:

A subset of $X$ is open if either it is empty or its complement is finite. If a subset of $X$ has infinite complement, it is open if and only if it is empty. But the empty set is not a neighborhood of any point.
Let's see it in another way. In order for $X$ to be Hausdorff, you need to find, for $x,y\in X$ with $x\ne y$ an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that $U\cap V=\emptyset$. Note that $U$ and $V$ are not empty.
Now, suppose they exist. Then $F=X\setminus U$ and $G=X\setminus V$ are finite. Therefore
$$
X=X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)=F\cup G
$$
is finite.

