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?
It is easy to construct filters which have no principal ultrafilters which extend them. For example, any filter extending the cofinite filter.
But in order to prove that there is any filter extending them to begin with, you need to use the axiom of choice.
Also, note that there is a difference between an ultrafilter on a set, and an ultrafilter on a Boolean algebra. It is true that you can reformulate the existence of ultrafilters as prime ideals on Boolean algebras, but there is something to be said there.
It might well be the case that your defined filter is an ultrafilter on the subalgebra of rays. But how does that help you finding an ultrafilter on the set $\Bbb Q$?