How to prove this closed formula for Cantor set? 
Let $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
Theorem: $$C_n=\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$$

I have tried to prove this assertion by induction on $n$, but to no avail. I am stuck at inductive step.
Please shed me some light to accomplish the proof. Thank you so much!

My attempt:
The formula is trivially true for $n=0$. Let it hold for $n$.
$$C_{n+1}=\frac{C_n}{3} \cup\left(\frac{2}{3}+\frac{C_n}{3}\right)$$
$$=\left(\frac{1}{3} \bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]\right) \cup \left(\frac{2}{3}+\frac{1}{3} \bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right] \right)$$
$$=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right) \cup \left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k+2.3^m}{3^{m+1}},\frac{2k+2.3^m+1}{3^{m+1}}\right]\right)$$
$$=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right) \cup \left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2(k+3^m)}{3^{m+1}},\frac{2(k+3^m)+1}{3^{m+1}}\right]\right)$$
$$=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right) \cup \left(\bigcap_{m=0}^{n}\bigcup_{k=3^m}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor+3^m}\left[\frac{2k}{3^{m+1}},\frac{2k+1}{3^{m+1}}\right]\right)$$
 A: Notice that 
$$\bigcap_{m=0}^{n+1}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]\right)\cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]\\=C_n\cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,\frac{1}{3^{n+1}}],\ldots,[\frac{3^{n+1}-1}{3^{n+1}},1]$.
$\bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]$ is just taking the even parts of the list together, notice that $C_{n+1}\subseteq \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]\cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $x\in C_n$, it had to be in the form $0.d_1d_2\ldots_3$ where $i\in \{1,2,\ldots, n-1\}$ implies $d_i\in\{0,2\}$, or in other words $x\in[0.d_1\ldots d_{n-1}0_3,0.d_1\ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n \cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\right]\subseteq C_{n+1}$
