How $\operatorname{cl}(A) \cap \operatorname{cl}(B) = \operatorname{cl}(A \cap B) \implies$ $X$ is the discrete space? I know that if $A,B \subset X$ then $\operatorname{cl}(A \cap B) \subset \operatorname{cl}(A) \cap \operatorname{cl}(B)$ for every topological space $X$, but how I use that $ \operatorname{cl}(A) \cap \operatorname{cl}(B) \subset \operatorname{cl}(A \cap B) $ to arrives that $X$ is the discrete topological space?
 A: Suppose $A \subseteq X$, and note that our hypothesis (by which I assume you mean that for every $A, B \subseteq X$ we have $\operatorname{cl}(A \cap B) = \operatorname{cl}(A) \cap \operatorname{cl}(B)$) implies
$$\emptyset = \operatorname{cl}(\emptyset) = \operatorname{cl}(A \cap A^\mathsf{c}) = \operatorname{cl}(A) \cap \operatorname{cl}(A^\mathsf{c})$$
and hence shows that $\operatorname{cl}(A)$ and $\operatorname{cl}(A^\mathsf{c})$ are disjoint.
On the other hand, in general we have
$$ X = \operatorname{cl}(X) = \operatorname{cl}(A \cup A^\mathsf{c}) = \operatorname{cl}(A) \cup \operatorname{cl}(A^\mathsf{c})$$
so we find that $\operatorname{cl}(A^\mathsf{c}) = \operatorname{cl}(A)^\mathsf{c}$. 
But now consider the fact that $A \subseteq \operatorname{cl}(A)$ and $A^\mathsf{c} \subseteq \operatorname{cl}(A^\mathsf{c}) = \operatorname{cl}(A)^\mathsf{c}$. The latter inclusion implies $\operatorname{cl}(A) \subseteq A$, so combined with the former inclusion we see that $\operatorname{cl}(A) = A$.
In other words, we have shown that an arbitrary set is closed, and hence $X$ is discrete. 
