Does every uncountable closed subset of $2^\mathbb N$ have cardinality $\mathfrak c$? Does every uncountable closed subset of $2^\mathbb N$ have cardinality $\mathfrak c$?
I think so: by applying a tree argument plus compactness, it suffices to show that given an uncountable closed set $A\subset 2^\mathbb N$ there exist disjoint uncountable closed sets $A_1, A_2 \subset A$. But is it true? How should I try to split $A$?
 A: You're on the right track. Call a point $a$ a condensation point of $A$ if every neighborhood of $a$ contains uncountably many points of $A.$ Show that any uncountable set $A$ has a condensation point. In fact, all but countably many points of $A$ are condensation points. So choose two condensation points and find disjoint closed neighborhoods of each, und so weiter.
A: I found a different solution.
We will show that if $F\subset 2^{\mathbb N}$ is closed and uncountable, there exists $F_1, F_2$ closed disjoint uncountable subsets of $F$ such that $F=F_1\cup F_2$ (this is stronger than what I asked)
Let $F\subset 2^\mathbb N$ be a uncountable closed set. For every natural number $n$ and for every $i \in \{0, 1\}$, let $U(n, i)=\{f \in 2^\mathbb N: f(n)=i\}$. Notice that this set is clopen.
There exists $i_0$ such that $U(0, i_0)\cap F$ is uncountable. Suppose we have chosen $i_n$ such that $F\cap \bigcap_{j=0}^n U(j, i_j)$ is uncountable. Pick $i_{n+1}$ such that$F\cap \bigcap_{j=0}^n U(j, i_j)\cap U(n+1, i_{n+1})$ is uncountable.
For every $n$, let $K_n=\bigcap_{j=0}^n U(j, i_j)$. Notice that for every $n$, $K_n$ is clopen, $F\cap K_n$ is uncountable and $F=(F\cap K_n)\cup (F\setminus K_n)$. Now it suffices to show that there exists $n$ such that $F\setminus K_n$ is uncountable. Notice that $\bigcap_{n \in \mathbb N}K_n=\bigcap_{n \in \mathbb N}U(j, i_j)=\{f\}$, where $f(n)=i_n$ for every $n$. Since $F$ is compact, $F\cap \bigcap_{n \in \mathbb N}=\{f\}$. Therefore:
$$\bigcup_{n \in \mathbb N}F\setminus K_n=F\setminus\bigcap_{n \in \mathbb N}K_n=F\setminus \{f\}$$
Since $F\setminus \{f\}$ is uncountable, there exists $n$ such that $F\setminus K_n$ is uncountable.

Now let's apply the tree argument. Let $T=\bigcup_{n \in \omega}2^n$ ($n=\{0, 1, \dots, n-1\}$, $0=\emptyset$).
Let $s^\frown t$ be the concatenation of $s$ and $t$.
Let $F_\emptyset=F$. Recursively, suppose we have defined $F_s$ (for some $s \in T$) as an uncountable closed subset of $2^\mathbb N$. By the previous lemma, let $F_{s^\frown(0)}$, $F_{s^\frown(1)}$ be two disjoint uncountable closed subsets of $F_s$.
Suppose $f \in 2^\omega$. Let $A_f=\{F_{s|n}: n \in \mathbb N\}$. By compactness, $\bigcap A_f$ is nonempty and by construction, if $f \neq g$ then $\bigcap A_f\cap \bigcap A_g=\emptyset$. For each $f \in 2^\mathbb N$, let $\phi(f) \in \bigcap A_f$. The function $\phi: 2^\mathbb N\rightarrow F$ is injective, therefore:
$$\mathfrak c=|2^\omega|\leq |F|$$
A: We can show by a Cantor-like construction that a non-empty complete metric space with no isolated points has a subspace $S$ homeomorphic to the Cantor set. For this Q if suffices that $|S|=\mathbb c.$  
If $T$ is an uncountable closed subset of  $\mathbb R$, let $U$ be the union of all open intervals $(a,b)$ such that $a,b\in \mathbb Q$ and such that $T\cap (a,b)$ is countable. Then $U$ is the largest open subset of $\mathbb R$ that has countable intersection with $T$. And  $T$ \ $U$ is a closed uncountable subspace of $\mathbb R$ with no isolated points. Since it is closed  in $\mathbb R$, it is a complete metric space. Therefore $|T$ \ $U|=\mathbb c.$
Now $2^{\mathbb N}$ is homeomorphic to the Cantor set $C$, and $C$  is closed in $\mathbb R$, so any $T\subset C$ which is closed in $C$ is also closed in $\mathbb R.$ So if $T\subset C$ is closed in $C$ and is uncountable then $|T|=\mathbb c.$
P.S. I am reminded of an old Q in American Mathematical Monthly: "A student asserted that every uncountable set of reals contains a closed uncountable subset. Is this true?" (By closed, we mean closed in the reals.)
