A set of real numbers without a perfect subset

Question

I am trying to solve this exercise from Jech's Set Theory (not an homework). I have two issues with it:

1. how can I define $a_0$ and $b_0$ (may I take them arbitrarily in $\mathbb R$?),
2. if there is a perfect subset $P$ of $A$, then there is $\alpha$ such that $P=P_\alpha$. From here I want to show that $P_\alpha$ contains at least one element in $B$ to get a contradiction but I am not sure how to do it.

Answer 1

For (1), $a_0$ can be indeed arbitrary, but $b_0$ is chosen in the method describe in the book: $b_0\in P_0\setminus\{a_0\}$.

For (2), assume that $P_α=P\subseteq A$ is perfect, in particular $P_α∩B=∅$, but either $b_α∈P_α∩B$, which is a contradiction, or $P_α\setminus \{a_ξ\mid ξ≤α\}=∅$, which is also a contradiction as $|P_α|=2^{\aleph_0}>|α|=|\{a_ξ\mid ξ≤α\}|$.