One way to describe the topology of the Baire space $\mathbb{B} = \omega^\omega$ is that the basic open sets are of the form $N_\eta = \left\{ f \in \omega^\omega \middle |\ \eta \subseteq f \right\}$ where $\eta \in \omega^{< \omega}$. Are the basic open sets closed?
I've seen it implied that they are, but it's not obvious to me that it would be the case.
Let $f \notin N_\eta$. So $\eta \nsubseteq f$, so there is some $n$ in the domain of $\eta$ such that $\eta(n) \neq f(n)$. Now create $\eta'$ to be the partial function $\{(n, f(n)\}$ of one element. Then $f \in N_{\eta'}$ and if $g \in N_{\eta'}$ then $g(n)= f(n) \neq \eta(n)$ hence $\eta \nsubseteq g$, so $N_{\eta'} \cap N_\eta = \emptyset$.
So any non-member has an open neighbourhood that completely misses it, so $N_\eta$ is closed.