How can we prove that there does not exist any metric space $4,5,6$ dense subsets? Let $(X,d)$ be a metric space. Which of the following is possible?
(A) $X$ has exactly $3$ dense subsets.
(B) $X$ has exactly $4$ dense subsets.
(C) $X$ has exactly $5$ dense subsets.
(D) $X$ has exactly $6$ dense subsets.
I think it's answer is (A) as if we consider $\mathbb R$ with usual metric then $\mathbb R$ contains three dense subsets $\mathbb Q$ ,$\mathbb Q^c$ and $\mathbb R$ itself.
Am I correct?Are there other metric spaces which makes (B),(C),(D) true? If not How can we prove that there does not exist any metric space $4,5,6$ dense subsets? 

In this answer I showed that only exactly 4 dense sets is possible: this can happen if almost all points of $X$ are isolated. Say there is a set $N$ of $n$ non-isolated points . Then $D\subset X$ is dense iff $D= (X\setminus N) \cup A$ where $A\subseteq N$. 
So there are as many dense subsets of$X$ as there are subsets of $N$, i.e. $2^n$. 
