Assuming the real line can be well ordered, a Bernstein set exists

Let ${\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}$ be the family of nonempty perfect subsets of $\mathbb{R}$. Every such set is closed, so it is determined by the set of rational open intervals disjoint from it; since rational intervals are countably coded, [L2] gives an injection ${\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}\to \mathbb{R}$. Conversely the map $a\mapsto [a,a+1]$ is an injection $\mathbb{R}\to {\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}$, because each interval $[a,a+1]$ is nonempty and perfect in the sense of Perfect subset of $\mathbb{R}$: closed with no isolated points. Hence [L3] gives ${\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}\approx \mathbb{R}$. Let $\kappa :=\mathrm{\mid }\mathbb{R}\mathrm{\mid }$. The given well-order of $\mathbb{R}$ makes $\mathbb{R}$ well-orderable, so [L3] gives $\mathrm{\mid }{\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}\mathrm{\mid }=\kappa$ and shows that $\kappa$ is a cardinal. The interval $[0,1]$ is a nonempty perfect subset of $\mathbb{R}$, so [L1] shows $\kappa$ is infinite; therefore [L4] gives $\mathrm{\mid }{\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}\times 2\mathrm{\mid }=\kappa$. Fix a bijection $e:\kappa \to {\mathcal{P}}_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}\times 2$.

For $\alpha \in \kappa$, write $e(\alpha )=(P_{\alpha },{\epsilon }_{\alpha })$ with ${\epsilon }_{\alpha }\in 2$. Suppose $z_{\beta }$ has been defined for every $\beta <\alpha$, and put $Z_{\alpha }:=\{z_{\beta }:\beta <\alpha \}$. By construction each new $z_{\beta }$ is chosen outside the earlier set $Z_{\beta }$, so the map $\beta \mapsto z_{\beta }$ is a bijection from $\alpha$ onto $Z_{\alpha }$. Hence $\mathrm{\mid }{Z}_{\alpha }\mathrm{\mid }=\mathrm{\mid }\alpha \mathrm{\mid }<\kappa$, because $\alpha \in \kappa$ and $\kappa$ is a cardinal. The set $P_{\alpha }$ is nonempty perfect, so [L1] and [L3] give $\mathrm{\mid }{P}_{\alpha }\mathrm{\mid }=\kappa$; therefore $P_{\alpha }\setminus Z_{\alpha }$ is nonempty. Let $z_{\alpha }$ be the $\prec$-least point of $P_{\alpha }\setminus Z_{\alpha }$. By transfinite recursion [L5], this defines a function $\alpha \mapsto z_{\alpha }$ on $\kappa$.

Put $B:=\{z_{\alpha }:\alpha \in \kappa \text{ and }{\epsilon }_{\alpha }=0\}$. Let $P$ be any nonempty perfect subset of $\mathbb{R}$. Since $e$ is onto, there exist $\alpha ,\beta \in \kappa$ with $e(\alpha )=(P,0)$ and $e(\beta )=(P,1)$. Step 2.1 gives $z_{\alpha }\in P$ and $z_{\beta }\in P$. Moreover $z_{\alpha }\in B$ by definition, while $z_{\beta }\mathrm{\notin }B$ because ${\epsilon }_{\beta }=1$. Thus every nonempty perfect subset of $\mathbb{R}$ meets both $B$ and $\mathbb{R}\setminus B$, so [F1] makes $B$ a Bernstein set.