Topological Conditions Equivalent to "Very Disconnected" Definition: Let $(X, \mathcal{T})$ be a topological space, where the set $X$ has more than one element. Suppose that for every pair of distinct elements $a, b \in X$, there exists a separation $(A,B)$ of $X$ such that $a \in A$ and $b \in B$. Then we say $(X, \mathcal{T})$ is very disconnected.
Is this condition (being "very disconnected") equivalent to another, well-known one?
The definition above is my own, but I suspect it is equivalent to some pre-existing notion (e.g., a $T_{n \frac{1}{2}}$ space for some $n$). Here are a few propositions that I have proved about v.d. spaces:


*

*Any very disconnected space is disconnected.

*Any discrete space is very disconnected.

*There are very disconnected spaces that are not discrete.

*If a space is very disconnected, then all singletons are closed.

*All singletons closed does not imply the space is very disconnected.
 A: You can define a relation where $x\sim y$ if there is no separation of the space $X$ between $x$ and $y$, or equivalently, each clopen set containing $x$ also contains $y$. It is very easy to check that this is an equivalence relation and that the equivalence class of $x$ is the intersection of all clopen neighborhoods. This class is called quasi-component of $x$.
This relation is coarser than the relation defining connectedness, so a quasi-component is a disjoint union of components. There are conditions under which the components and the quasi-components coincide, for example when $X$ is compact Hausdorff, or when there are only finitely many quasi-components. It also holds if the components are open (which is the case for locally connected spaces or spaces with only a finite number of components.)
An example of a totally disconnected space, i.e. all components are singletons, where the quasi-components are not the singletons is the sequence 
$$\left\{\frac1n\mid n\in\Bbb N\right\}\cup\{0,0'\}$$
converging to two distinct zeros, where the neighborhood base of $0$ is given by the intervals $[0,\epsilon),\ \epsilon>0$ and similarly for $0'$. Then each $0$ is a component but the quasi-component of $0$ is $\{0,0'\}$
Edit: @Chris Eagle points out that a space where the quasi-components are the singletons is called totally separated.
