Partition $\mathfrak c$-dense set to $\mathfrak c$- many dense set

Question

A subset of $A\subset\mathbb R$ is called $\mathfrak c$-dense if $ |A\cap I|=\mathfrak c$ for any none open interval $I\subset\mathbb R.$ Then, there is a partition for $A$ to continuum many dense set. Here is my idea: Let $\{I_{\xi}\colon \ \xi<\mathfrak c\}$ be enumeration of all open interval subset $R$ and since $A$ is $\mathfrak c$-dense, so we have $A_{\xi}= A\cap I_{\xi}$ and $|A_{\xi}|=\mathfrak c.$ Put $\mathcal A=\{A_\xi\colon \ \xi<\mathfrak c\}.$ Let $\{\langle r_\xi, A_\xi\rangle\colon \ \xi<\mathfrak c\}$ be an enumeration of $A\times \mathcal A$ with $\mathfrak c$-many repetitions, since $|A\times \mathcal A|=\mathfrak c$. By induction on $\xi<\mathfrak c$, choose a sequence $$ x_\xi\in A_\xi\setminus \{x_\zeta\colon \ \zeta<\xi\}$$Now, for $r\in A$, let $D_r=\{x_\xi\colon r_\xi=r\}$. Notice that the points $x_\xi$ are all distinct. so the sets $D_r$'s are pairwise disjoint and it is to check they are dense. Now, if $$A\setminus\bigcup D_r=B.$$ So we can add the set $B$ to one of the sets $D_r$ and this finishes the proof. Please let me know if there is some thing wrong. Thank you in advance.