Show that every uncountable and closed subset of the complete separable metric space contains a homeomorphic subset with the Cantor set. Show that each uncountable and closed subset of the complete separable metric space contains a homeomorphic subset with the Cantor set.
How to show it?
 A: First of all note that if $X$ is a Polish space (completely metrizable and separable space), and $F\subseteq X$ is closed, then $F$ is also Polish (usually with a different metric). The result you want can be shown to hold with a combination of the following two classical results:

Theorem 1: Let $X$ be a nonempty perfect Polish space (perfect means without isolated points), then there is an embedding of the cantor space $C$ into $X$.
Theorem 2 (Cantor-Bendixson): Let $X$ be a Polish space. Then $X$ can be written as a disjoint union $P\cup S$, where $S$ is countable open and $P$ is perfect.

Those results and much more material concerning Polish spaces can be found in the book by Kechris Classical Descriptive Set Theory, I'll write down the proofs for completeness, which I took from the aforementioned book.
To prove Theorem 1 we need the concept of a Cantor scheme on a space $X$. A family of subsets $(A_s)_{s\in 2^{<\Bbb N}}$ of $X$ is called a Cantor scheme if:

*

*$A_{s0}\cap A_{s1}=\varnothing$ for all $s\in 2^{<\Bbb N}$

*$A_{si}\subseteq A_s$ for all $s\in 2^{<\Bbb N}$ and all $i\in\{0,1\}$
Intuitively this is a way to split up the space that mimics the middle-third construction of the Cantor set.
Proof of Theorem 1: Let $X$ be a perfect Polish space. We will construct a Cantor scheme $(A_s)_{s\in 2^{<\Bbb N}}$ on $X$ satisfying:

*

*$A_s$ is open and nonempty for every $s\in 2^{<\Bbb N}$.

*$\mathrm{diam}(A_s)\leq 2^{-|s|}$.

*$\overline{A_{si}}\subseteq A_s$ for all $s\in 2^{\Bbb N}$ and all $i\in\{0,1\}$
Then, identifying $C$ with the product space $\{0,1\}^\Bbb N$, we define a map $f\colon C\to X$ by letting $f(c)=\bigcap_{n\in\Bbb N} \overline{A_{c|n}}$, which is a singleton because it is a decreasing intersection of closed sets with diameter going to $0$ in a complete metric space. Checking that $f$ is continuous and injective is easy, and any continuous injection from a compact space to an Hausdorff one is an embedding. We only need to construct such a Cantor scheme, which is easily done by induction on $|s|$. For $s=\varnothing$ let $A_\varnothing$ be an arbitrary nonempty open set in $X$. Given $A_s$ we can construct $A_{s1}$ and $A_{s0}$ by picking distinct elements of $A_s$ ($A_s$ cannot be a singleton since $X$ is perfect) and letting $A_{s1}$ and $A_{s0}$ be small enough balls around those two elements.
Proof of Theorem 2: Let $$P=\{x\in X\mid \text{every open nbhd of $x$ is uncountable}\}, $$ and let $S=X\setminus P$. $C$ is open countable because if $\{B_n\}_{n\in\Bbb N}$ is a countable basis of the topology of $X$, then $C$ is the union of those $B_n$ which are countable. To see that $P$ is perfect let $x\in P$. Then every nbhd of $x$ is uncountable, but then it must contain uncountably many points in $P$, since $X\setminus P$ is uncountable, so $x$ is not isolated in $P$ and $P$ is perfect.
