This is the part of proof in Rudin PMA p.41
Let $P(\subset \mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,…$.
Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m \in P$ such that (i) $\overline {V_{n+1}}$ $\subset V_n$ and (ii) $x_n \notin \overline {V_{n+1}}$ and (iii) $V_{n+1} \cap P ≠ \emptyset$. Then form a sequence $\{V_n \subset \mathbb{R}^k | n\in \omega \}$.
Here, Axiom Of dependent choice is used.
I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)
I want a proof without AC.
Help..
Best Answer
The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.
In the induction step, instead of taking "some $x_m\in P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.
Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.
Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.