"The elements in the Cantor set are the end points of all the intervals in $E_n$..." This is your mistake. This isn't true. In fact, written in ternary expansion, the elements of the Cantor set are precisely those elements in $[0,1]$ with a ternary expansion consisting of $0$'s and $2$'s (where we note $0.01=0.00\bar{2}\in\mathcal{C}$, but $0.0101\notin\mathcal{C}$, for example). Using this fact, it isn't hard to show that $\frac{1}{4}\in\mathcal{C}$ but $1/4$ is not an endpoint of any interval.
Let $C_0=[0,1.$ For $n\geq 0$ we have $C_n=\cup F_n$ where $F_n$ is a finite set of pairwise-disjoint closed intervals, each of positive length. And the measure $m(C_n)$ is $\sum_{f\in F_n}m(f).$
For $n\geq 1,$ at stage $n$ we remove an open interval $I_f$ from each $f\in F_{n-1},$ with $m(I_f)=\alpha \cdot m(f).$ So the measure $$m( C_n)=\sum_{f\in F_{n-1}}(1-\alpha)m(f)=(1-\alpha)m(C_{n-1}).$$ So by induction on $n$ we have $m(C_n)=(1-\alpha)^n\cdot m(C_0)=(1-\alpha)^n. $
Therefore $m(\cap_{n\geq 0}C_n)\leq \inf_{n\geq 0}m(C_n)=0.$
For $n\geq 0$ let $L(n)$ be the length of the longest member of $F_n.$ Each $f\in F_n$ has a central open piece $I_f$ of length $\alpha \cdot m(f)$ removed at stage $n+1,$ resulting in $2$ members $f',f''$ of $F_{n+1},$ each of length $\frac {1-\alpha}{2}m(f).$ Therefore by induction on $n$ we have $$L(n)= ((1-\alpha)/2)^{-n}L(0)<2^{-n}L(0)=2^{-n}.$$ So if $K$ is an interval in $[0,1]$ of length $m(K)>0,$ then for any $n$ large enough that $2^{-n}<K$ we have $K\not \subset C_n,$ and a fortiori we have $K \not \subset \cap_{n\geq 0}C_n.$
The open intervals removed at stage $n\geq 1$ are disjoint from each other and are disjoint from the open intervals removed at any previous stage. So the family $I$ of all open intervals that are removed is a pair-wise disjoint family, so $$\sum_{i\in I}m(i)=m(\cup_{i\in I}\;i)=m(\cup I)=m([0,1]\backslash \cap_{n\geq 0}C_n)=1.$$
BTW some students erroneously suppose that $C=\cap_{n\geq 0}C_n$ consists entirely of the end-points of the members of $I$. But all limit-points of all convergent sequences of those end-points also belong to $C$.
Best Answer
If you take two sets which are constructed by removing centrally situated open intervals you have a for each iteration a homeomorphism between the two constructions, e.g. a piecewise linear map defined on the partition of [0,1]. These homeomorphism have a limit function which provides a bijection on the two Cantor-like sets.