Can we have a countable region in a perfect set? I am reading a proof in Baby Rudin that perfect sets in $\mathbb R^k$ are uncountable. The proof seems to assume it is first countable and show that it contradicts the fact that if a collection of compact sets have non empty finite intersections then the entire intersection is nonempty. 

However, is it possible for an open neighborhood of a point in an uncountable set, say $x_0,$ to intersect only a countably infinite number of points? Because then it seems as if the same proof can be applied to this subset only.
Consider the set $\{0\}\cup\{1/n : n \in \mathbb{N}\} \cup\{x :1 \leq x\leq 2, x \in \mathbb{R}\}$ which is uncountable and any ball of radius less than one around zero will only have countably many points and is an open neighborhood. Notice that it has isolated points so it is not perfect.

Also consider $\{x : 0 \leq x \leq 1, x \in \mathbb{Q}\} \cup\{x : 1 \leq x \leq 2, x \in \mathbb{R}\}$ which is uncountable, any of the rational numbers on the interval $[0,1]$ can have a ball of sufficiently small radius to contain only a countable number of points and it has no isolated points, but it is not closed so it is not perfect.
Consider the set $\{0\}\cup\{1/n : n \in \mathbb{N}\} \cup\{x :1 \leq x\leq 2, x \in \mathbb{R}\}$ which is uncountable and any ball of radius less than one around zero will only have countably many points and is an open neighborhood. Notice that it has isolated points so it is not perfect.

Also consider $\{x : 0 \leq x \leq 1, x \in \mathbb{Q}\} \cup\{x : 1 \leq x \leq 2, x \in \mathbb{R}\}$ which is uncountable, any of the rational numbers on the interval $[0,1]$ can have a ball of sufficiently small radius to contain only a countable number of points and it has no isolated points, but it is not closed so it is not perfect.