# Equivalence of $T_1$ axiom definition

I already know the following definition for $$T_1$$ axiom:

$$(1)$$ Let $$X$$ a topological space. We say $$X$$ satisties $$T_{1}$$ axiom if all the finite subset of $$X$$ are closed.

Now, I want to prove that definition is equivalent to the following definition:

$$(2)$$ Let $$X$$ a topological space. We say $$X$$ satisfies $$T_{1}$$ axiom if for all distincts $$x,y\in X$$, there are neighborhoods $$x\in U_{x}, y\in U_{y}$$ in $$X$$ such that $$x\notin U_{y}$$ and $$y\notin U_{x}$$.

To prove $$(1)\Rightarrow(2)$$, we know that $$A:=\{x,y\}$$ is closed in $$X$$ for all distincts $$x,y\in X$$. Suppose that for all neighborhood $$U_{x}\ni x,U_{y}\ni y$$ we have $$x\in U_{y}$$ and $$y\in U_{x}$$. How can I get a contradiction with the fact that $$A^{c}$$ is open?

To prove $$(2)\Rightarrow (1)$$, we can show that $$\{x,y\}$$ is closed in $$X$$ for all distincts $$x,y\in X$$, and that is enough, but I don't know how to do that.

Can someone give me some hints?

$$(1)\implies(2)$$ Take $$x,y\in X$$, with $$x\neq y$$. Since $$\{x\}$$ is closed, $$X\setminus\{x\}$$ is open and, since $$y\neq x$$, $$y\in X\setminus\{x\}$$. There you have it: $$X\setminus\{x\}$$ is a neighborhhod of $$y$$ no which $$x$$ doesn't belong.
$$(2)\implies(1)$$ If $$x\in X$$, then, for each $$y\in X\setminus\{x\}$$, lete $$A_y$$ be an open set such that $$x\notin A_y$$. Then $$\bigcup_{y\in X\setminus\{x\}}A_y=X\setminus\{x\}$$. This proves that $$X\setminus\{x\}$$ is open. So, $$\{x\}$$ is closed. Since this occurs for every singleton…