I've read the definition of a topological Hausdorff space (two distinct points have disjoints neighbourhoods) and of a disconnected space (it is the union of two disjoint nonempty open sets)
Now I think an equivalence definition of a topological space being Hausdorff would be this:
Prop 1) A topological space $X$ is Hausdorff iff every two point subset $\{x,y\} \subseteq X$ (with $x \neq y$) is a disconnected subspace of $X$.
Proof. $\Rightarrow$) $X$ is Hausdorff. If $X = \emptyset$ or a singleton it is valid. Else, for any $\{x,y\} \subseteq X$ with $x\neq y$ we take the disjoints (open) neighbourhoods $x \in E$, $y \in V$, and so by the subspace topology $\{x\}$ and $\{y\}$ are open sets of $\{x,y\}$, so $\{ \{x\}, \{y\} \}$ is a disconnection, and $\{x,y\}$ is disconnected.
$\Leftarrow$) If every $\{x,y\} \subseteq X$ with $x\neq y$ is disconnected, then there are (open) neighbourhoods $E,V$ such that $x \in E$, $y \in V$ and $E \cap V = \emptyset$. This are disjoint neighbourhoods and is valid $\forall x,y \in X$, so $X$ is Hausdorff.
Is this ok? If it is, it's strange to me that it isn't usually mentioned in the wikipedia/textbooks (or I haven't seen it at least).
Edit: It seems Proposition 1 is false. I'll try to show the following
Prop 2) A topological space $X$ is $T_1$ iff every two point subset $\{x,y\} \subseteq X$ (with $x \neq y$) is a disconnected subspace of $X$.
Proof. $\Rightarrow$) $X$ is $T_1$ so for every $x,y \in X$, $x$ has a neighbourhood $E$ not containing $y$, and $y$ has a neighbourhood $V$ not containing $x$. So $\{x\}$ and $\{y\}$ are open in $\{x,y\}$, non empty and disjoint, so $\{x,y\}$ are disconnected.
$\Leftarrow$) Every $\{x,y\}$ is disconnected, let $E,V$ be a disconnection, then there exists open sets $E',V'$ of $X$ such that $E = E' \cap \{x,y\} = \{x\}$ and $V = V' \cap \{x,y\} = \{y\}$. This $E', V'$ are neighbourhood of each point not containing the other, and is valid for every $x, y \in X$, so $X$ is $T_1$.
Is this ok now?
This isn't quite right. The problem is that in the reverse direction, you don't know that $E$ and $V$ are open in $X$; you only know that they are open in the subset topology on $\{x,y\}$. This means that there are open sets $E',V'\subseteq X$ such that $E=E'\cap \{x,y\}$ and $V=V'\cap\{x,y\}$. But now you don't know that $E'$ and $V'$ are disjoint, because they might intersect at points of $X$ other than $x$ and $y$.
In fact, your condition is equivalent to the $T_1$ axiom, not the Hausdorff axiom. Can you prove this?