I believe that Shelah's model gives a ZFC theorem of relative consistency results:
$ZFC \vdash Con(ZFC + \text{there is a strongly inaccessible cardinal})\rightarrow Con(ZF+DC+ \text{all sets are measurable} + \text{there exists a set without the property of Baire}).$
Now I wonder if $$ZF+? \nvdash [\text{there exists a set without the property of Baire}\rightarrow \text{there exists a non-measurable set}],$$ where ? is a suitable assumption. So ? is not full AC.
I try to show that DC is suitable. By completeness theorem, assume the meta theory of first order logic is ZFC set theory, we just find a model of ZF+DC where all sets are measurable while there exists a set without the property of Baire.
Now, this meta ZFC can be regarded as a theory in the meta-meta first order logic so completeness holds. But I don't know how ZFC can directly provide such a model, unless we refine this meta theory as ZFC+Con(ZFC+IC) so that Shelah's result applies. Is this modification harmless?
It seems ? can't be DC+Con(ZFC+IC). It may be harder to find a model of ZF+DC+Con(ZFC+IC). If we try the other method by directly weakening the choice in Shelah's result, there seems to be a problem. In the construction of generic extension, at least DC is used, for ctm to choose a generic G and for Levy collapse maybe some "global DC" is used so that it's already enough to construct a non-measurable set. Further, in the argument for generic extensions satisfying certain properties some kind of choice may also be used, but I don't know.
Of course, it's good if there is any direct result in descriptive set theory distinguishing these two regular properties, or Shelah's method could be modified in a forcing-free setting.
I’m not exactly sure what you’re asking here. If you’re asking about what level of choice is necessary in the metatheory to do relative consistency proofs involving the axiom of choice, the answer is none. $\DeclareMathOperator{Con}{Con}$
Keep in mind that $(ZF + ?? \nvdash $ there is a set without the property of Baire $\to$ there exists a non-measurable set$)$ is exactly equivalent to $\Con(ZF + ?? + $ there is a set without property of Baire $+$ all sets are measurable$)$. So I’m not exactly sure why you decided to rephrase the problem in this way.
You say you want to work in the metatheory of ZFC and unconditionally prove a statement of the form $\Con(ZF + ?? + $ there is a set without property of Baire $+$ all sets are measurable$)$. Assuming ZFC is consistent, this is impossible by Gödel’s second incompleteness theorem, since you would then be able to prove the clearly weaker statement $\Con(ZF)$, which is (ZFC provably) equivalent to $\Con(ZFC)$.
In fact, we know that $\Con(ZFC + $ there exists an inaccessible cardinal$)$ and $\Con(ZF + DC + $ all sets are measurable$)$ are equivalent. See here. Depending on how exactly one defines “Lebesgue measurable” in the absence of choice, it is possible for $ZF + $ all sets are measurable to be equiconsistent with ZF. If you choose a “bad” definition of measurable, it follows from the claim that $\mathbb{R}$ is the countable union of countable sets, which is relatively consistent with ZF.