I understand the definition of $T_D$ but fail to see how it constitutes a separation. $T_0$ has one open set that contains one point of a pair of points but not the other. So this is clearly a separation condition. The same is the case with $T_1$ where both points have an open set that contains the point but not the other. In $T_2$ these separating open sets are additionally disjoint. So why does the exemption of a point from an open set separate points?
Edit1: the definition of $T_D$ is either that $\downarrow x - \{x\}$ is closed or that for open $U$ that contains $x$ $U-\{x\}$ is open too. E.g. the space $X=\{a,b,c\}$ with topology $\tau=\{\emptyset,\{a\},\{a,b\},\{a,c\},X\}$ is $T_D$. The points $a$ and $b$ are $T_0$ separated and the points $b$ and $c$ are $T_1$ separated.
Edit2: $\downarrow x$ is the downward closure: $\downarrow x = \{a| a \leq x\}$ or without an order on the points the intersection of all closed sets containing $x$. And $\downarrow x - \{x\}$ is the closure minus the point of interest. Somehow the $T_D$ property seems to be a single point separation criterion.