Consider the following separation property:
For any countable disjoint subsets $A, B\subseteq X$ of a $T_1$ space $X$ such that $A\cap \overline{B} = \overline{A}\cap B = \emptyset$ there exist disjoint open $U, V$ with $A\subseteq U, B\subseteq V$.
Say that a space $X$ has the countable separation property if it satisfies the above. If there exists another name for it then I'd like to know it, I came up with this name.
$$T_3\text{-space} \implies \text{countable separation property}\implies T_2\text{-space}$$
We say that a space is $T_{2\frac{1}{2}}$ if any two distinct points have disjoint closed neighbourhoods. There are examples of Hausdorff spaces which don't have countable separation property, but can be either $T_{2\frac{1}{2}}$ or not $T_{2\frac{1}{2}}$.
Question:
Does there exist spaces which aren't $T_3$ but have countable separation property? If yes, does such space need to be $T_{2\frac{1}{2}}$? Or maybe can't be $T_{2\frac{1}{2}}$?
Yes, there are such spaces.
A topological space $X$ is called a P-space if every $G_\delta$ subset of $X$ is open.
First of all let’s check that any Hausdorff P-space has the countable separation property, so let $X$ be a P-space and take two disjoint countable sets $A=\{a_i\}$ and $B=\{b_j\}$. By Hausdorfness for every $i,j$ there are disjoint open sets $a_i\in A_{ij}^1$, $b_j\in A_{ij}^2$. Using that $X$ is a P-space we now have disjoint open sets $$A_i^1=\bigcap_j A_{ij}^1,\quad A_i^2=\bigcup_j A_{ij}^2$$ separating $a_i$ from $B$. Using that $X$ is a P-space we finally obtain open sets $$\bigcup_iA_i^1,\quad \bigcap_iA_i^2$$ separating $A$ and $B$.
So all we need now is an Hausdorff P-space which is not regular, for an example see the beginning of section 3 of this paper and the references therein