Regular space has a sequence of disjoint open sets Exercise Suppose $X$ is a regular topological space. Suppose that $B \subset X$ is infinite. Prove that there exists a sequence of open sets $\{U_i\}_{i \in \mathbb{N}}$ satisfying
(1) $U_j \cap U_k = \varnothing$ for all $j,k \in \mathbb{N}$
(2)  $U_i \cap B \neq \varnothing$ for all $i \in \mathbb{N}$
I need help figuring out this proof. I have a couple of suggestions.

*

*Since $U_1, U_2, U_3, \dots$ needs to be a sequence, we need to work with a countably infinite subset of $B$. On the other hand, a countably infinite subset of $B$ might prevent from having $U_i \cap B \neq \varnothing$


*We design a sequence of $U_1,U_2,U_3, \dots$ (somehow?) and then take the closure $\overline{U_i}$ and use the regularity to have that $U_i \subset \overline{U_i}$ with $\overline{U_i} \cap U_k = \varnothing \implies U_i \cap U_k = \varnothing$
There seems to be a few moving pieces going on in this exercise. Can anyone offer any advice on a good starting point for this problem?

Since this is not true for the indiscrete topology, I assume that regularity includes T1. Then proceed as follows:
(1) If $X$ is T2, $B$ an infnite subset of $X$, there is an open $U \subseteq X$ such that $U \cap B \neq \emptyset$ and $B \setminus U$ is infinite. 
(2) By regularity, you can improve (1) by ... and $B \setminus \overline{U}$ is infinite. 
(3) Proof the claim by induction using (2).
