Solution 1:

They don’t have to be open: a neighbourhood base can even consist of compact neighbourhoods only, say.

The only requirements for a neighbourhood base at $x$ are : all the sets in it must be neighbourhoods of $x$ ( they are: Wilansky shows there is an open $G$ with $x \in G \subseteq B(x)$) and every other (open) neighbourhood of $x$ contains one of the $B(x)$ as a subset ( and he shows that too).