 Solution

The answer is no. Let $X_0=\mathbb{N}$, regarded as one point of the space of infinite subsets of the natural numbers, and putThe displayed union makes $O$ an open set in the Ellentuck topology. On the other hand, every basic neighbourhood $[s,A]$ of $X_0$ contains a different infinite set: keep $s$ and remove one element from its infinite remaining tail. This other point belongs to $O$. Therefore $X_0$ is in the closure of $O$ but not in $O$, soThe fact that basic neighbourhoods are clopen sets does not make arbitrary unions of them closed.