How is $T_D$ a separation axiom 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.
 A: As the title of the reference given by Dave L. Renfro suggests, $T_D$ is logically strictly between $T_0$ and $T_1$. In that very direct sense, $T_D$ must convey some separation information about spaces. I will try to elaborate on that below.
First, since $T_D$ is strictly between $T_0$ and $T_1$, note that every $T_D$ space is a $T_0$ space (but not the other way around). So, if $T_0$ guarantees some sort of separation, so will $T_D$.
To see that every $T_D$ space is  $T_0$, assume for contradiction that there is a space that is  $T_D$ which further has two distinct points $x,y$ that have exactly the same neighborhoods (i.e. not a $T_0$ space). Take an open set $U$, such that $x,y \in U$. Then, in particular $x \in U$. By the assumption that the space is $T_D$, $U- \{x\}$ is open. But then $U- \{x\}$ is an open that contains $y$ and not $x$, which contradicts our assumption.
The above argument answers the question:

why does the exemption of a point from an open set separate points?

To further understand the difference between $T_D$ and $T_0$, I think its instructive to compare similar $T_0$ spaces, where one fails to have the $T_D$ property.
Start by considering the space $S$ over the natural numbers with opens of the form $U_k = \{n \in \mathbb{N} | n \geq k \}$ for every natural $k$ (together with the empty set). Basically, every ray going to the right over the naturals is an open set here.
This space is $T_0$. Every two distinct natural numbers $a, b$ have different neighborhoods. Moreover, it is $T_D$. Note that the specialization order of this topology is just $\leq$ over the naturals. Take some natural number $m$. Then $\downarrow m - \{m\}$ is just all the naturals strictly smaller than $m$. Such set is closed (it is the complement of $U_m$, which is open by construction). So the space is $T_D$. Further, its not $T_1$.
Now consider the very similar $S'$, a space over the set of natural number plus an extra point $p$, $\mathbb{N} \cup \{p\}$, and where opens are now  $U_k = \{n \in \mathbb{N} | n \geq k \} \cup \{p\}$ for every natural $k$.
$S'$ is still $T_0$. Take $p$ and any natural number $n$, and the open $U_{n+1}$ contains $p$ but not $n$.
However, $S'$ is not $T_D$. The set $\downarrow p - \{p\}$ is $\mathbb{N}$, but $\mathbb{N}$ is not closed as $\{p\}$ is not open.
Adding the point $p$ clearly "messed something up."  Intuitively, the set of points in $S$ are more separable than the set of points in $S'$. The $T_D$ property, which $S$ has and $S'$ lacks, explains the difference. For extra fun, observe that only $S$ is an Alexandrov topology.
I hope this helps.
See this link for further observations along these (and many more) lines:
https://projects.lsv.ens-cachan.fr/topology/?page_id=2626
