There are opens $U,V$ such that $a\in U-V, b\in V-U$ $\iff $ every finite subset is closed Definition: the $T_1$ axiom says that given $2$ distinc points $a,b\in X$, there will exist opens $U$ and $V$ such that $a\in U-V$ and $b\in V-U$.
I need to show that a topological space has the $T_1$ property $\iff$ every finite subset is closed.
I'm trying to do the following: I need to prove that every finite subset is closed, that is, the complementar is open. But I don't even know if the complementar is finite or infinite. So... Maybe I need to suppose that there exists a finite subset which is open. I don't know also how this would help. I guess that, since it's a finite set of points, I can take balls as in the $T_1$ axiom and maybe take their intersection and arrive at something.
Could somebody help me?

First let $X$ be $T_1$ we show that every finite subset is closed. As Marco suggested it suffices to prove that every singleton set is closed. Now assume this is not the case, say there exists $\{a\}\subseteq X$ which is not closed, this means $X-\{a\}$ is not open, thus by definition there exists $b\in X-\{a\}$ such that $X-\{a\}$ does not contain any open neighbourhood of $b$. This means every open neighbourhood of $b$ intersects $X-(X-\{a\})=\{a\}$, which contradicts the assumption $X$ is $T_1$ (because from the definition of $T_1$ we can find at least one open neighbourhood of $b$ which does not contain $a$). 
The converse is straight forward: If every finite subset is closed, this means every singleton set is closed. Thus for distinct $a,b\in X$, choose $U=X-\{b\}$ and $V=X-\{a\}$, we are done.
