Q:

Condition on closures implies discreteness of topology.

I'm supposed to prove that if $(X, \tau)$ is a topological space such that $\overline{A \cap B} = \overline{A} \cap \overline{B}$ for all $A,B \subset X$, then $\tau$ is the discrete topology.
My first ideas were to try proving that $\{x\}$ is open for all $x \in X$, or proving that $A$ is closed, whichever $A \subset X$ is. I would show some work, but I couldn't manage anything so far. I mean, looking at extreme cases using $A$ or $B$ equal to $\{x\}, X \setminus \{x\}, X$ or $\varnothing$ didn't gave me anything useful.
Can someone give me ideas? Thanks.

A:

For some $x \in X$ suppose that $X - \{x\}$ is not closed.  Then clearly $\overline{X - \{x\}} = X$.  Then we have that 
$$
\emptyset \;\; =\;\; \overline{ \{x\} \cap (X - \{x\})} \;\; =\;\; \overline{\{x\}} \cap \overline{X - \{x\}} \;\; =\;\; \overline{\{x\}} \cap X
$$
but this is a contradiction.  Therefore we must have that $X - \{x\}$ is closed.

