Theorem 17.6 in Munkres, thank you @Ted. So the closure of $\bar{A}$ is equal to the union of $A$ itself and the set of its limit points.
Thus, in the case of these dense subsets of $X$ denoted $(x_n)_{n=1}^{\infty}$, we know that their closures must be equal to the unions of the $x_n$s themselves as well as their corresponding limit points, resulting in those open balls.
I think that's right?