In the article on ultrafilters, Wikipedia claims that
In ZF without the axiom of choice, it is possible that every ultrafilter is principal.{see p.316, [Halbeisen, L.J.] "Combinatorial Set Theory", Springer 2012}
I assume this means that "All ultrafilters are principal" is consistent with ZF, i.e. that there exist models of ZF in which this statement holds. This is also confirmed by this question.
Now I don't know a lot about model theory and also no do not necessarily need to know the details, I was just surprised because I thought one could construct examples of non-principal ultrafilters. For example, consider the sub-Boolean-algebra $\mathcal{B}$ of $\mathcal{P}(\mathbb{Q})$ generated by the sets of the form
$$ \lbrace x \in \mathbb{Q} \mid a < x \rbrace, \lbrace x \in \mathbb{Q} \mid a > x \rbrace $$
for $a \in \mathbb{Q}$. Then surely the set
$$ \lbrace \lbrace x \in \mathbb{Q} \mid 0 < x < a \rbrace \mid a > 0 \rbrace $$
generates a non-principal ultrafilter
$$ \mathcal{U} = \lbrace U \in \mathcal{B} \mid \exists a > 0: \lbrace x \in \mathbb{Q} \mid 0 < x < a \rbrace \subseteq U \rbrace $$
in $\mathcal{B}$, and this can be concluded without using the axiom of choice?