Is there a simple reason why uncountable compact sets of real numbers have cardinality continuum?
I know that this is immediate from the Cantor-Bendixon Theorem, but I wonder whether this consequence in its own can be proved in a simpler/shorter manner. Perhaps building a Cantor set inside the compact set is one answer to my question, but even this is not so trivial if you did not see similar things before (many small assertions need to be proved in such a case). I wonder if there are better approaches.
The question is relevant for a survey paper I write, where going into the CB Theorem (just for the mentioned purpose) may distract the attention of the reader off the main thread of the paper.
Update: As of writing this update, all suggested proofs are variations of constructing a Cantor set inside our compact set, or the proof of CB Theorem using condensation points. It seems to me that, unless a simpler proof is suggested, this means that the easiest proof to explain is based on the following lemma.
Lemma. Every uncountable compact real set has uncountable intersection with two disjoint closed intervals.
The lemma can be proved, e.g., by showing that there are at least two condensation points (points all of whose neighborhoods are uncountable). This follows from the reason mentioned by @fedja: Otherwise, our compact set is countable. Another way to prove this is provided by @Fedor Petrov.
As of now, it seems that there is no simpler argument.