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.



A simpler proof that compact sets have cardinality continuum?
I think "the" conceptual reason why an uncountable compact subset of $K\subset \mathbb{R}$ has the cardinality of the continuum is that it has a "canonical" map onto a space homeomorphic to $[0,1]$.


The complement of $K$ is a union of open intervals: a left ray, a right ray and at most countably many bounded ones.
The canonical map is the quotient map to the space obtained by gluing the end points of these bounded intervals in the complement
(it is easy to see that the equivalence classes of the equivalence relation thus obtained are at most countable).


This answers the question in the sense that this is a "simple reason why...". I am not saying that this is the simplest proof, as one still has to argue that the quotient space is homeomorphic to $[0,1]$ (a choice of such a homeo is certainly not canonical). There are various ways to do this.




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**Edit:** below I replace a previous argument (for $K/{\sim}\simeq [0,1]$) by a more conceptual one.


Recall that $K$ is assumed uncountable and that the equivalence relation $\sim$ we defined on it has countable fibers. Denote $X=K/{\sim}$.
Endow it with the quotient topology and quotient order. 
The following is easy.


Observation: $X$ is a connected, separable, compact linearly ordered space which is not a singleton.


(separablity and compactness are inherited from $K$, connectedness follows from the definition of $\sim$ and $X$ is not a singleton is by the fact that the equivalence classes are countable - this is where we use that $K$ is uncountable).


We are left to prove the following.


Proposition: Every space $X$ saisfying the properties above is homeomorphic to $[0,1]$.


Fix a countable dense subset $A\subset X$. Assume as you may that $\min X$ and $\max X$ are not in $A$.
The proof of the proposition consists of the combination of the following facts:


1. $X$ is an order completion of $A$.
2. $A$ is order isomorphic to $\mathbb{Q}$.
3. The order completion of $\mathbb{Q}$ is the two points compactification of $\mathbb{R}$.


Facts 1 and 3 are easy (details could be found in <https://en.m.wikipedia.org/wiki/Dedekind-MacNeille_completion>).
In fact, fact 3 could be regarded as the definition of $\mathbb{R}$. For fact 2, observe that $A$ is a countable dense linear order with no min and max. It is a classical fact (due to Cantor) that every two models of this theory are isomorphic, see <https://en.m.wikipedia.org/wiki/Dense_order>.
The proof is given in <https://en.m.wikipedia.org/wiki/Back-and-forth_method>. It is a first-course-in-set-theory-exercise (but don't give it in the exam).


