I know that any nulldimensional Polish space $X$ has a clopen basis $-$ because it is nulldimensional and a countable basis $-$ because it is Polish. But do all nulldimensional Polish spaces have a basis that is countable and clopen?
I couldn't find any answers online, so any help would be appreciated!
Yes. One way to see this is that every zero-dimensional Polish space $X$ embeds as a subspace of the Cantor space $2^\omega$ (see Theorem I.7.8 on p. 38 of Kechris's Classical Descriptive Set Theory). Now $2^\omega$ has a countable basis of clopen sets, and the intersection of each of these basic clopen sets with $X$ provides a countable basis of clopen sets for $X$.
Here's an alternative approach which might seem more elementary:
It is a general topological lemma that if a space $X$ is second-countable, then any basis has a countable subset which is still a basis. In particular, given a basis $\mathcal{B}$ of clopen sets for a zero-dimensional Polish space, $\mathcal{B}$ has a countable subset which is still a basis of clopen sets.
For a proof of the topological lemma, see bof's answer here.
Since this answer is rather terse and is on a closed question with a delete vote, I'll rewrite the proof here with a few more details. Let $\mathcal{B}' = \{B_n\mid n\in \omega\}$ be a countable basis for $X$. Let $\mathcal{B}$ be an arbitrary basis for $X$. For each pair $m,n\in \omega$, if there is any $U\in \mathcal{B}$ with $B_m\subseteq U\subseteq B_n$, pick one and call it $U_{m,n}$. Otherwise let $U_{m,n}$ be an arbitrary element of $\mathcal{B}$. I claim that the countable set $\{U_{m,n}\mid m,n\in \omega\}\subseteq \mathcal{B}$ is a basis for $X$. Let $x\in V\subseteq X$ with $V$ open. Since $\mathcal{B}'$ is a basis, pick $n$ such that $x\in B_n\subseteq V$. Since $\mathcal{B}$ is a basis, pick $U\in \mathcal{B}$ such that $x\in U\subseteq B_n\subseteq V$. Since $\mathcal{B}'$ is a basis, pick $m$ such that $x\in B_m\subseteq U\subseteq B_n\subseteq V$. Now $U$ may not be in our countable basis, but we have $x\in B_m\subseteq U_{m,n}\subseteq B_n\subseteq V$, so $x\in U_{m,n}\subseteq V$, and we're done.
A proof of the natural generalization of the lemma with "countable" replaced by "cardinality $\kappa$" is given in Brian M. Scott's answer here.