$(X,\tau)$ be a finite topological space , then is it Regular if ( and only if ) it has a basis which forms a partition of $X$? Let $X$ be a finite set and $\tau $ be a topology on $X$ such that $(X,\tau)$ is a Regular space ( https://en.wikipedia.org/wiki/Regular_space  , not assuming Hausdorff property) ; is it true that  $(X,\tau)$ has a basis which forms a partition of $X$ ? Is the converse true i.e. if $X$ is a finite set and $\tau$ is a topolgy on $X$ such that $(X,\tau)$ has a basis which forms a partition of $X$ , then is $(X,\tau)$ a Regular space ?
 A: The answer is "yes".
Let $\tau$ be a topology on a finite set $X$ and suppose that $(X, \tau)$ is a regular space. I claim that every closed subset $F$ of $X$ is also open. Indeed, for each $x \notin F$, there exists an open set $U_x$ such that $x \notin U_x$ and $F \subseteq U_x$ (as you can see, I just need a weaker property that regularity).
Then the set $U = \bigcap_{x \notin F} U_x$ is open (as a finite intersection of open sets) and is equal to $F$ by construction.
It follows that $\tau$ is a finite Boolean algebra and its atoms form a partition of $X$ and a basis for $\tau$.
Conversely, suppose that $\tau$ has a basis which forms a partition of $X$. Then the elements of this basis are clopen and it follows that every open set is clopen. Finally, if $F$ is closed and $x \notin F$, then $F$ is clopen and thus $X - F$ is also clopen. Therefore $x \in X-F$, and the disjoint open sets $X-F$ and $F$ separate $x$ from $F$.
A: Yes. If $(X, \tau)$ is a finite topological space, then the collection of its minimal neighbourhoods is a base. For $x$ in $X$, write $U_x = \bigcap \{U : x \in U \in \tau\}$ for the minimal neighbourhoods and $\mathscr{U} = \{U_x x\in X\}$ for the base. These sets have the wonderful property that if a point $x$ is contained in any open set $U$ then automatically $U_x \subseteq U$. This "minimality" is used repeatedly in the proof of one direction below.
Suppose $(X, \tau)$ is regular and finite. For $x \in X$ and $F = U_x^c$ we can get $U$ and $V$ open and disjoint such that $U \ni x$ and $V \supseteq F$ (by regularity). But then $U_x \subseteq U$ by minimality, and $U \cap V = \varnothing$ implies $V \subseteq U^c$, and thus $V \subseteq U_x^c = F$. So $V = F$; that is, $F$ is also open, hence $U_x$ is also closed. So we have shown that in a finite regular space, $\mathscr{U}$ consists of clopen sets. 
If $U_x$ and $U_y$ be two distinct base sets, then one of the following is true: $x \not\in U_y$ or $y \not\in U_x$. For if both were false, i.e. if $x \in U_y$ and $y \in U_x$, then we would have $U_x \subseteq U_y$ and $U_y \subseteq U_x$ and hence equality. So w.l.o.g. we assume $y \not\in U_x$. Then $y \in U_x^c$, which is open, so by minimality $U_y \subseteq U_x^c$, which is equivalent to $U_x \cap U_y = \varnothing$. Therefore, the sets in $\mathscr{U}$ are pairwise disjoint. It follows that $\mathscr{U}$ is a partition of $X$, since $\bigcup_{x\in X} U_x = X$.
Showing that a partition space is regular is much easier, and it is true for arbitrary spaces. For in a partition space, every open set is closed and every closed set is open. Therefore, if $F$ is closed and $x$ is a point not in $F$, then $F$ is also open, and $(F, F^c)$ is the pair we seek to establish regularity. $\blacksquare$
Note. Spaces $(X, \tau)$ admitting a base of clopen sets are sometimes called zero-dimensional.
