$\mathbb R$ as continuum many of pairwise disjoint of Bernstein sets The set $B\subset\mathbb R$ is called Bernstein set if neither $B$ nor $\mathbb R\setminus B$ contains any perfect sets.
Theorem: $\mathbb R$ can be written as continuum many of pairwise disjoint Bernstein sets.
Proof:  Let $\mathcal{P}$ be a family of all perfect subsets of $\mathbb R$. Then  $|\mathcal{P}\times\mathbb R|=\mathfrak c , $ so we can choose an enumeration $\{\langle P_\xi,y_\xi\rangle\colon\xi<\mathfrak c\}$ of $\mathcal{P} \times \mathbb R.$
We will construct, by induction on $\xi<\mathfrak c,$ a
sequence $\{x_\xi\colon \xi<\mathfrak c\}$ such that
$$x_\xi\in P_\xi\setminus\{x_\zeta\colon \zeta<\xi\}$$
Since each $x_\lambda\neq x_\xi$ for all $\lambda<\xi<\mathfrak c$, we can define $f$ on $\{x_\xi\colon \xi<\mathfrak c\}$ such that $f(x_\xi)=y_\xi$ and $f(x)=0$ otherwise.
Claim : $f^{-1}(r)$ is Bernstein set for each $r\in\mathbb R.$
Indeed, for each $r\in\mathbb R$ and $P\in\mathcal P$ there is an $\xi<\mathfrak c$ such that $\langle r,P\rangle = \langle r_\xi, P_\xi \rangle$,  then $x_\xi\in f^{-1}(r_\xi)\cap P_\xi=f^{-1}(r) \cap P\neq \emptyset.$  Also the same true for the complement for $f^{-1}(r)$ (as $f^{-1}(t)$ where $t\neq r$).  Then $f^{-1} (r)$ is Bernstein for all $r\in\mathbb R$. It is clear that $$ \mathbb R=\bigcup_{r\in\mathbb R} f^{-1} (r)$$ as we need.
I have no probelm with this proof at all. My question is Can we have the same result without  define function like that? I mean just start to construct the by transfinite induction. I would love to see completely different approache to get the same result. I am hoping for theos who have deeply understanding for transfinite induction to share their proofs.
 A: First, and least important, your construction is by recursion, not by induction: induction is a proof technique.
Next, the argument doesn’t really use the function $f$ at all: you could just as well define your Bernstein sets by letting $B_r=\{x_\xi:y_\xi=r\}$ for each $r\in\Bbb R$.
Finally, your construction does not ensure that the union of your Bernstein sets is all of $\Bbb R$; it could, for instance, be $\Bbb R\setminus\Bbb Q$ if you happened to choose irrational numbers for all of your points $x_\xi$. If you want them to be a partition of $\Bbb R$, you have to be a bit more careful in your recursion.
Enumerate $\Bbb R=\{r_\xi:\xi<\mathfrak{c}\}$. At stage $\eta$, when you’ve already chosen $x_\xi$ for all $\xi<\eta$, let $\mu=\min\big\{\zeta<\mathfrak{c}:r_\zeta\in P_\eta\setminus\{x_\xi:\xi<\eta\}\big\}$, and let $x_\eta=r_\mu$. This is just your construction with a little extra care taken in choosing which point of $P_\eta\setminus\{x_\xi:\xi<\eta\}$ to use as $x_\eta$, so it does give you pairwise disjoint Bernstein sets $B_r=\{x_\xi:y_\xi=r\}$ for $r\in\Bbb R$.
Suppose that $\Bbb R\setminus\bigcup_{r\in\Bbb R}B_r\ne\varnothing$, let $r_\nu\in\Bbb R\setminus\bigcup_{r\in\Bbb R}B_r$ and let $X=\{\xi<\mathfrak{c}:r_\nu\in P_\xi\}$; then $|X|=\mathfrak{c}$, and $x_\xi\in P_\xi$ for each $\xi\in X$. For each $\xi\in X$ there is a $\mu(\xi)<\mathfrak{c}$ such that $x_\xi=r_{\mu(\xi)}$, and the points $x_\xi$ are distinct, so the map $\mu$ is injective. Let $M=\{\mu(\xi):\xi\in X\}$; $|M|=\mathfrak{c}$, so we can let $\theta=\min\{\mu\in M:\nu\le\mu\}$. Note that $\theta=\mu(\eta)$ for some $\eta\in X$, and $x_\eta=r_\theta$.
By hypothesis $r_\nu\notin\bigcup_{r\in\Bbb R}B_r$, so there is no $\xi<\mathfrak{c}$ such that $x_\xi=r_\nu$, and therefore $\theta>\nu$. But this is impossible: $r_\nu\in P_\eta\setminus\{x_\xi:\xi<\eta\}$, and $\nu<\theta$, so at stage $\eta$, when we chose $x_\eta$, $\theta$ was not the least element of $\big\{\zeta<\mathfrak{c}:r_\zeta\in P_\eta\setminus\{x_\xi:\xi<\eta\}\big\}$, and we did not set $x_\eta=r_\theta$.
Added: Here are a few links to answers of mine containing constructions by transfinite recursion, and one to an old paper of mine that does so; I’ve arranged them very roughly in ascending order of complexity. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 [PDF; see Theorem $\bf{5}$].
