Question: I want to prove that every closed subset of $\mathbb R$ is finite or countable or continuum.

I know that for arbitrary subset we can not make similar statements - because of continuum hypothesis. 

It looks like I need some machinery to prove mentioned fact. I feel I need somehow characterize all closed subsets of $\mathbb{R}$. For open subsets I know that they are union of disjoint intervals. For closed I do not know anything like that. Intution tells me that some deep unknown for me fact should be used here.

Answer: 
Assume that $F$, the closed set, is uncountable. Then, we can show that there exist two disjoint closed intervals $I\_0=[a\_0,b\_0]$ and $I\_1=[a\_1,b\_1]$ such that
$$
\lvert F\cap I\_0\rvert>\aleph\_0 \,\,\,\&\,\,\,\lvert F\cap I\_1\rvert>\aleph\_0.
$$ 
In order to to this, we first pick an $x\in F$, such that 
$\lvert(x-\varepsilon,x+\varepsilon)\cap F\rvert>\aleph\_0$, for all $\varepsilon>0$. If no such $x$ existed, then we could cover $F$ with countably many open intervals (since $F$ is separable), each of which intersects $F$ at countably many points, and hence $F$ would be countable. Clearly, this point $x$ can not be unique, for otherwise $F\!\smallsetminus\!\{x\}$ would similarly be covered by countably many open intervals, each intersecting $F$ at countably many points. If $x<y$ are two such points, and $3d=y-x$, then we can set
$$
I\_0=[x-d,x+d]\quad\text{and}\quad I\_1=[y-d,y+d].
$$

Repeating the same argument, with $F\cap I\_0$ and $F\cap I\_1$ in the place of $F$, we next find four disjoint closed intervals $I\_{0,0}$, $I\_{0,1}$, $I\_{1,0}$ and $I\_{1,1}$, such that $I\_{0,0},\, I\_{0,1}\subset I\_0$ and
$I\_{1,0},\, I\_{1,1}\subset I\_1$, and
$$
\lvert F\cap I\_{i,j}\rvert>\aleph\_0, \quad i,j=0,1.
$$ 
Continuing this procedure, for every sequence $j\in 2^{\mathbb N}=\{0,1\}^{\mathbb N}$ we have a shrinking to zero sequence of closed intervals
$$
I\_{j(1)}\supset I\_{j(1),j(2)}\supset I\_{j(1),j(2),j(3)}\supset I\_{j(1),\ldots,j(n)}\supset\cdots
$$ 
each of which has as an intersection a singleton $\{x\_j\}$, and clearly $x\_j\in F$, as $F$ is closed, for every $j\in 2^{\mathbb N}$. But, by construction of these closed intervals, if $j\ne j'$, then $x\_j\ne x\_{j'}$. Thus we have constructed an injection $\varphi :2^{\mathbb N}\to F$, and hence
$$
\lvert F\rvert\ge 2^{\aleph\_0}.
$$
Let $C$ be the set. First, without loss of generality:

* $C$ is nowhere dense.
* $C\subseteq [0,1]$

Show that if $C$ is uncountable, there's always an open interval outside of $C$ which divides $C$ into 2 uncountable halves.

If you repeat this process in both halves and so on, compactness implies there's a point in every possible branching, so there's continuum many of them.