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? 

Hint: Consider the set $X = A\cup D$ with $A=\{0\}$ and $D=\left\{\frac1n: n\geq 1\right\}$ equipped with the metric from ${\Bbb R}$. Show that the only dense subsets are $A\cup D$ and $D$ (so there are precisely two dense subsets).
Now, this example is constructed using a discrete set $D$ with precisely one accumulation point $0$. Try now to  construct sets using a discrete set and having 2, 3,... accumulation points and draw conclusions from this.
