Singleton sets closed in T_1 and Hausdorff spaces

[complexnumber]

If $$(X,\tau)$$ is either a $$T_1$$ space or Hausdorff space then for any $$x \in X$$ the singleton set $$\{ x \}$$ is closed.

Why is this the case? I can't see the reason from the definitions of the spaces.

Definition:
Let $$(X,\tau)$$ be a topological space and let $$x,y \in X$$ be any two
distinct points, if there exists any two open sets $$A,B \in \tau$$
such that $$x \in A$$ but $$x \notin B$$ and $$y \in B$$ but $$y \notin A$$,
then $$(X, \tau)$$ is a $$T_1$$ space.

Definition:
A topological space $$(X, \tau)$$ is Hausdorff if for any
$$x,y \in X$$, $$x \ne y$$, $$\exists \text{ neighborhoods } U \ni x$$ and
$$V \ni y$$ such that $$U \cap V = \varnothing$$.

 

[Dick]

{x} is closed if the complement of {x} is open. Try to show X/{x} is open using the definitions. That's not so hard, is it?