Choice gives a Bernstein set with no perfect-set, Baire or measure regularity

By F1 and A1 fix a well-order of $\mathbb{R}$ and apply F2 to obtain Bernstein B. Its complement is Bernstein too, since F3's two intersection conditions are symmetric. Neither side contains a nonempty perfect P: such P must also meet the other side by F3, contrary to containment.

We prove the category avoidance needed below. Given a nonempty open interval J and a sequence of closed nowhere dense F_n, choose the least rational bounded interval I_empty of length less than one with closure inside $J\setminus F_0$. Given I_s at depth n, choose the least coded pair of rational nonempty child intervals with disjoint closures inside $I_s\setminus F_{n+1}$ and lengths less than $1\mathrm{/}(n+2)$. Such pairs exist: the complement of the closed nowhere dense set has a nonempty open piece in I_s; that piece contains two separated rational intervals by F8. F9's recursion, with defaults outside valid histories, constructs all levels, and the preceding existence proves defaults unused. Put $K={\bigcap }_n{\bigcup }_{\mathrm{\mid }s\mathrm{\mid }=n}\overline{I_s}$. Each level is closed (a finite union), so K is closed.

Each binary branch gives nested nonempty bounded closed intervals with lengths tending to zero by F9; F7 supplies its unique point, inside J and outside every F_n. Hence K is nonempty. A point of K has a unique interval at each level by disjoint sibling closures and determines a branch, so these are exactly K's points. For x in K and $ϵ>0$, take a level interval on its branch of length less than $ϵ$. Follow the opposite child at the next level and then always the left child. F7 supplies a different point of K in that same parent interval, by disjoint child closures; its distance from x is less than $ϵ$. Thus K has no isolated point and is a nonempty perfect set by F3.

No Bernstein set is meagre: otherwise close its nowhere dense covering witnesses and apply step 2.1 in (0,1) to obtain a nonempty perfect set missing it, contrary to F3. If B had the Baire property, choose open U and closed nowhere dense F_n covering $B\mathrm{△}U$. If U were empty B would be meagre, already excluded. Otherwise choose an interval J inside U and use step 2.1 to find nonempty perfect $K\subseteq J\setminus {\bigcup }_n{F}_n\subseteq B$, contradicting step 1.1. The same argument applies to the complement. Countable real sets are meagre, since singletons are closed nowhere dense and an enumeration (padded for finite sets) supplies witnesses; hence neither side is countable. By F6 and A1 every Borel set has the Baire property, so neither side is Borel.

AC supplies countable choice: a choice function on the range of a sequence of nonempty sets, composed with that sequence, chooses its terms. Therefore F4 and F5 apply to B and to its Bernstein complement from step 1.1. F5 gives nonmeasurability of both. F4 gives inner measure zero and the exact outer measure value for B in every specified interval, regardless of its endpoint convention. These conclude all assertions. QED.