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
$$
\lvert F\rvert\ge 2^{\aleph\_0}.
$$
Here's an outline of how you can prove that an uncountable closed subset $X$ of $\mathbb R$ has the cardinal of the continuum. I guess you know that it's enough to find an injective (one-to-one) mapping from $\mathbb R$ to $X$. (There's an obvious injection in the other direction, and then the Cantor-Bernstein theorem gives you a bijection.)

1. For any uncountable set $X\subseteq\mathbb R$, there is a real number $a$ such that the sets $\{x\in X:x\lt a\}$ and $\{x\in X:x\gt a\}$ are both uncountable.
2. In fact, for any uncountable set $X\subseteq\mathbb R$, there is a number $a\in X$ such that the sets $\{x\in X:x\lt a\}$ and $\{x\in X:x\gt a\}$ are both uncountable.
3. For any uncountable set $X\subseteq\mathbb R$, there is an order-embedding of $\mathbb Q$ into $X$; that is, a function $f:\mathbb Q\to X$ such that $x\lt y\implies f(x)\lt f(y)$.
4. If $X$ is an uncountable **closed** subset of $\mathbb R$, then the function $f$ from the previous step can to extended to an order-embedding $\hat f:\mathbb R\to X$. (Namely, for each irrational $x$ we can define $\hat f(x)=\sup\{f(q):q\in\mathbb Q\}$.)