Is every set an open set?

I'm reading Rudin's "Principles of mathematical analysis", definition 2.18 stated that:

"A neighborhood of $p$ is a set $N_{r}(p)$ consisting of all $q$ such that$d(p,q)<r$, for some $r>0$."

This does not require that $q$ is a different point from $p$.

"A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N\subset E$."

Then every point $p \in E$ is an interior point because we can always have a neighborhood containing only $p$. Thus every set will be open since every point can be an interior point.

However, he later stated that "A nonempty finite set is not open" in an example right below. Is he just forgot to add the constraint that a neighborhood can not contain only one point or I'm missing something here ?