This question is about some equivalents of the Hahn-Banach theorem in $\textsf{ZF}$ set theory. As far as I know, the definitive reference for this sort of thing is Howard & Rubin's Consequences of the Axiom of Choice, which I refer to below. (The Hahn-Banach theorem is "Form 52" in the book.)
The starting point of my questions about this topic is the following equivalent of Hahn-Banach:
[52D] Let $\mathcal B_0$ be a subalgebra of a Boolean algebra $\mathcal B$, and let $m_0$ be a real-valued finitely additive probability measure defined on $\mathcal B_0$. Then there is a real-valued finitely additive probability measure $m$ that is an extension of $m_0$ from $\mathcal B_0$ to $\mathcal B$, and the range of $m$ is contained within the closed convex hull of $m_0$.
The first question I had is:
To what extent are abstract Boolean algebras necessary in [52D]? In particular, if we require only that [52D] hold for Boolean algebras of subsets, as in [52?] below, does the equivalence with Hahn-Banach still hold?
In case you are tempted to appeal to Stone's representation theorem here, note that that theorem is stronger than Hahn-Banach, so we are not free to invoke it.
There are known measure-theoretic equivalents of Hahn-Banach that don't require abstract Boolean algebras. For instance:
[52C] For every nonempty set $X$ and every proper ideal $\mathcal I$ over the powerset of $X$, there is a real-valued finitely additive probability measure $m$ defined on every subset of $X$ such that $m(I)=0$ for every $I \in \mathcal I$.
Now, [52C] is clearly implied by the conjunction of
[A] For every nonempty set $X$ and every proper ideal $\mathcal I$ over the powerset of $X$, there is a real-valued finitely additive probability measure $m$ defined on $\mathcal A(\mathcal I)$ such that $m(I)=0$ for every $I \in \mathcal I$, where $\mathcal A(\mathcal I)$ is the algebra of subsets of $X$ generated by $\mathcal I$
and
[52?] For every set $X$, every algebra $\mathcal A_0$ of subsets of $X$, and every real-valued finitely additive probability measure $m_0$ on $\mathcal A_0$, there is a real-valued finitely additive probability measure $m$ that is an extension of $m_0$ from $\mathcal A_0$ to the powerset of $X$.
But it seems to me that [A] is a theorem of $\textsf{ZF}$. Indeed, every set in $\mathcal A(\mathcal I)$ is of the form $$\bigcup_{i=1}^n\bigcap_{j=1}^{m_i}A_{ij},$$ where every $A_{ij}$ is in $\mathcal I$ or its complement is. Thus, setting $m(I)=0$ for all $I \in \mathcal I$ uniquely determines a 0-1 valued probability measure on $\mathcal A(\mathcal I)$.
If that's right, then [52?] implies [52C]. And clearly [52D] implies [52?]. So [52?] is actually equivalent to Hahn-Banach, even though it looks quite a bit weaker than [52D]. Final questions:
Is this reasoning correct? If so, is there a reference for this result (it seems very likely that it's been noticed before, if true)?
Your first question is a bit tricky to answer, but in principle, the answer is no. We can arrange a model of $\sf ZF$ in which there is a proper class of "bad examples" to Hahn–Banach.
But one can also talk about $X$ as "$X$ is the least $V_\alpha$ such that there is a counterexample to Hahn–Banach of rank $\alpha$, or $X=\{\varnothing\}$". Then of course, it is enough to verify that there are no counterexamples of Hahn–Banach with underlying set contained in $X$. But this is the kind of reasoning that makes people hate mathematicians, so let's agree it is a formalist eye-rolling approach, and it misses the point, so we do not need to talk about it anymore.
To your second question, it might be worth noting several things:
The Hahn–Banach theorem talks about extending linear functionals. One can always get a subspace that admits a non-zero linear functional, e.g. one-dimensional subspaces. It's the extensions where the problems start to crop up.
Even the stronger principle, $\sf BPI$ (which is equivalent to Stone's representation theorem), can be stated as prime filters on abstract Boolean algebras, or the extensions of filters to ultrafilters (or ideals on power sets to maximal ideals).
The way to move between these two is simply to note that we can take a quotient of a power set by an ideal, get a Boolean algebra, and any ideal in that algebra extends our ideal.
My point here is that again the trick is the extension bit of the process.
Finally, the reasoning seems correct, but not for very interested reasons, as I pointed out above. I'm not sure whether this was put in print, but I'd start by looking at Eric S. Schechter's book "Handbook of Analysis and its Foundations" where there are a lot of discussion around Hahn–Banach.