I'm looking for the weakest separation axiom, which gives the following property:
Let $A$ be neighborhood of the point $x$. Then there exists another neighborhood $B$ of $x$, such $\overline{B}\subset A$.
I thought that $T_{3}$ would suffice, but I was able to get only closed set $C$ such $$x\in C\subset A $$
I could take $B=Int(C)$, but then I do not see that $B$ will be neighborhood of $x$, i.e. $x\in B$.
If no separation axiom per se, then in what space this property would hold? The more abstract, the better.
Your condition is equivalent to being regular.
Suppose $X$ is regular. Let $A$ be a neighborhood of $x$ and suppose without loss of generality that $A$ is open. Then regularity lets us choose disjoint open sets $U \ni x$ and $V \supset A^c$. That is, $U \subset V^c$ where $V$ is closed, so $\overline{U} \subset V^c \subset A$. Thus $B = U$ is the desired neighborhood of $x$.
Conversely, suppose $X$ has your property. Let $x \in X$ and let $F$ be a closed set not containing $x$. Then $A = F^c$ is a neighborhood of $x$. Suppose $B$ is a neighborhood of $x$ with $\overline{B} \subset A$. Set $U = B^\circ$, the interior of $B$, and $V = (\overline{B})^c$. Then $U,V$ are open and disjoint, $x \in U$, and $F \subset V$ since $V^c = \overline{B} \subset A = F^c$.