@DavidGao: Oh right, good point. I was just thinking of condition 1.

@DavidGao: I think it depends on the ideal. If the ideal is all sets of cardinality $<\!\lambda$ for some $\lambda \leq \kappa$, then the answer is still no, for essentially the same reason. But here's a silly example where the answer is yes: let $u$ be an ultrafilter on $\mathcal P(\omega)/(\mathrm{fin})$, and let $I$ be the corresponding ideal. Then $\mathcal P(\omega)/I$ is trivial, and you can get a section like what you want. Off the top of my head, I don't have any examples of ideals where you get a nice section for interesting reasons.

@BenjaminSteinberg: Yes, I agree. I expect that any proof of this theorem needs to make use of the topology on $\beta \mathbb N$, either openly or sneakily. That's what makes it a good example of what the OP is asking for (unless it's dismissed as not "real algebra" and more set theory).

There is an important semigroup operation $+$ defined on the set of ultrafilters on $\mathbb N$. It is an important and nontrivial theorem that there are non-principal ultrafilters $u$ for which $u+u = u$. The statement of this theorem, and all of the relevant definitions, can be expressed without any reference to any topology on the set of ultrafilters. But every proof of the theorem that I know involves topology in a nontrivial way. (I'm putting this as a comment, rather than an answer, because I imagine it might stretch too far what you mean by an "algebraic theorem.")

@AlessandroDellaCorte: I agree -- done.

@EmilJeřábek: I was assuming a standard Borel measure. But, looking again at the question, perhaps I shouldn't be?

Nice answer -- +1. Do you know whether your lower bound improves on the one in the question? In other words, do you know whether it is consistent that RR fails, but still every set surjects onto $\omega$?

Do you consider a "stronger" topology to have more open sets, or fewer? (I think there's some ambiguity in the way people use this word. Analysts tend to mean more open sets, I think, but I feel like I've heard it the other way too.)

Very nice answer, Joel! Let me add that a variant of this question, Is there a dense subposet with the upwards-finite property?, is more subtle. The answer is no for some posets and yes for others, and it seems a subtle question where exactly the dividing line lies. For example, it's independent of ZFC whether every ccc poset has this property.

The oldest still-open problem in set theory post-dates Hilbert's list, so it's not the answer to this question. But if you're interested, here's a link to a blog post (written by Asaf Karagila, whom you may recognize from MO) explaining the problem and some of its history: karagila.org/2014/on-the-partition-principle

(You probably know this already, but ) if there is a rigid planar curve, it must have unbounded curvature. Any self-entwined curve with constant curvature is locally homeomorphic to $\mathbb Q \times \mathbb R$, and therefore not rigid. By the way, do you have a non-planar example of a rigid curve?

For a specific example, let $B$ be a Bernstein subset of the reals, and let $A = \{ (x,-x) :\, x \in B\}$. This is a closed subset of the Sorgenfrey plane. I can see how to write it as a $G_\delta$ set (so it's not a counterexample to what Dan Ma's blog claims about the Sorgenfrey plane being perfect). But I cannot see how to show that $A$ is a regular $G_\delta$ set in the sense of the question.