A simpler proof that compact sets have cardinality continuum? 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: Let $[m,M]$ be the minimal segment containing our set $K$. Consider the set of such $x$ that $K\cap [m,x]$ is countable. It obviously contains its maximal element $u<M$. Analogously, consider the minimal $v$ for which $K\cap [v, M]$ is countable. We have $m<u<v<M$. Choose points $u<A<B<v$ and consider two uncountable compact subsets $K_0=K\cap [m,A]$ ,  $K_1=K\cap [B,M]$. Proceed this, constructing $K_{01}$ etc. For each infinite sequence of 0's and 1's $p_1p_2\dots$ we take a common point of all compact sets $K_{p_1\dots p_n}$. So, continuum many points in $K$ are found.
A: 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.

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:


*

*$X$ is an order completion of $A$.

*$A$ is order isomorphic to $\mathbb{Q}$.

*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).
A: Let $\ C\ $ be an uncountable compact subspace of $\ \mathbf R.\ $ Let W be the family of all open (in $\ C\ $) countable sets, and $\ V := \bigcup W.\ $ Family $\ W\ $ admits a countable covering of $\ V.\ $ Thus, the union $\ V\ $ is countable. Now, let's replace $\ C\ $ by $\ C\setminus V.\ $ The new $\ C\ $ is still uncountable. Furthermore, every nonempty open subset of $\ C\ $ is uncountable.
Let $\ K\ $ be the linearly ordered set obtained from $\ C\ $ by identifying every two $\ a\ b \in C\ $ such that $\ (a;b)\cap C=\emptyset.\ $ Set $\ K\ $ is still uncountable (there are only countably many points $\ x\in\ C\ $ which belong to classes which have more than one point). The topological space canonically associated with $\ K\ $ has the minimal and maximal point, and it is easy to check that $\ K\ $ has Dedekind property (for an interval) induced by Dedekind property of $\ \mathbf R.\ $ Thus $\ K\ $ is a connected Hausdorff compact space.
(There is a Urysohn function of $\ K\ $ onto a non-trivial interval).
A: You can make a few shortcuts, but it will still be the same proof in the end. For instance, you can consider the tree of all closed dyadic subintervals of some interval containing the set, after which (modulo the trivial lemma that boundary points are countable and a slightly less trivial lemma that continuum minus countable is continuum) the question reduces to the following:
You grow a tree and at each step you can either branch into two links or just add one. You want to show that the number of paths is either countable or continuum. Now take a sequence of 0's and 1's and play the usual game: the total number of paths starting from a vertex is countable if it is countable for both children. So, assume that the number at the root is uncountable. Then it is uncountable for at least one child. If it is uncountable for both children, you use the first number in your sequence to make a choice and forget that number. Otherwise pass to the child that is uncountable and repeat. If your choice is unique all the way down, you get one path and a countable union of countable sets (possibilities to go sideways), so the root must be countable, which is a contradiction. So, you'll eventually reach a fork with 2 uncountable children where you finally can use the first number to make your choice and so on.
The only advantage of this argument is that it is running straight from the two ideas I try to push down the students' throats anyway every time I teach an introductory analysis course: bisection and the nested interval lemma, so I am in a position to say: "You see, we are doing the very same stuff again and again!". It is very close to the bisection proof of the existence of the maximal value in spirit. Otherwise, it is more or less equivalent to any other approach, really.  
