Walter Rudin's mathematical analysis: theorem 2.43. Why proof can't work under the perfect set is uncountable. I found several discussions about this theorem, like this one. I understand the proof adopts contradiction by assuming the perfect set $P$ is countable.
My question is if the assumption is $P$ is uncountable, the proof seems remains the same, i.e., the $P$ can't be uncountable either. In other words, I think whatever the assumption is, we can draw the contradiction in any way.
I don't understand in which way the uncountable condition could solve the contradiction in the proof.

The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $\cap F$ is dense in $P.$
Suppose $P$ is a non-empty closed subset of $\Bbb R^n.$ Let $P$ inherit  the usual metric from $\Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $\Bbb R^n.$ Then $F=\{P \setminus \{x\}: x\in P\}$ is a non-empty countable family of dense open subsets of the space $P,$ so $\cap F=\emptyset$ is dense in $P,$ which is absurd.
(If $P$ were not assumed to be perfect then not all members of $F$ could  be assumed to be dense in $P.$)
Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.
