Show that the Cantor set is perfect i.e. $C=C'$ Let $C\subseteq \mathbb{R}$ be the classical Cantor set, show that $C=C'$.
So this is what i've done so far:
Take $x\in C$  $ \Rightarrow x \in C_{k}\text{  }\forall k$ 
$\Rightarrow \forall k \text{ , } x\in I_{k}(x)=\left [ a_{k},b_{k} \right ] \subset C_{k}$ 
where $I_{k}(x)$ is one of the disjoint intervals which union is $C_{k}$ and which containts $x$ (note that it is unique).
Also note that $\left [ a_{k+1},b_{k+1} \right ] \subset \left [ a_{k},b_{k} \right ] \text{ } \forall k\text{ and   }a_{k},b_{k}\in C \forall k$
\begin{equation}{(*)}
\Rightarrow \text{  } \bigcap_{n=1}^{\infty }I_{n}(x)=x
\end{equation}
\begin{equation}{(**)}
\Rightarrow \forall \epsilon >0 \text{   } \exists \text{     } a_{k(\epsilon)} \text{  s.t.  } |x-a_{k(\epsilon)}|<\epsilon
\end{equation}
\begin{equation}{(***)}
\therefore \text{    } B_{\epsilon}(x) \cap C \neq \emptyset
\end{equation}
$\therefore  \text{   } x \in C'  $ 
and since $x$ is arbitrary $C=C'$.
So what I'm no sure is if the "jump" from * to ** is correct, and also the "jump" from ** to *.
Should I add something to make it clearer? or change the argument, or is it totally incorrect?
 A: A proof that two sets are equal usually consists of two parts: one containment and the reverse containment. What you wrote proves $C'\subset C$. The containment $C'\subseteq C$ follows from $C$ being the intersection of closed sets. 

"jump" from * to ** 

Correct, but perhaps insufficiently detailed. You may want to invoke the fact that the intersection of closed nested intervals $[a_k,b_k]$ is $[\sup a_k, \inf b_k]$.
There's a problem  at the very end: you need $(B_\epsilon(x)\setminus \{x\})\cap C$ to be nonempty, not $ B_\epsilon(x) \cap C$. 

the "jump" from ** to *. 

Do you mean that the implication $*\implies **$ can be reversed? Not as stated. In general, don't do this when learning to write proofs. Rather, write out two separate parts as suggested above.

I would make the proof more quantitative, hence more precise: every interval in $C_k$ has length $3^{-k}$. Since $a_k\le x\le b_k$, it follows that $|x-a_k|\le 3^{-k}$ and $|x-b_k|\le 3^{-k}$. Given $\epsilon>0$, you can find $k$ such that $3^{-k}<\epsilon$. And at least one of $a_k,b_k$ is different from $x$.
