Proving any $x \in [0,1]$ belongs to infinitely many $S^{k}_{n}:=[\frac{k-1}{2^n},\frac{k}{2^n}]$ I am trying to prove the following:

Define $S^{k}_{n}:=[\frac{k-1}{2^n},\frac{k}{2^n}]$
i) Given any $x \in [0,1]$, then $x$ belongs to infinitely many $S^{k}_{n}$
ii) Any $x \in [0,1]$ also belongs to the complement of infinitely many such $S^{k}_{n}$

Ideas
It is suggested to use binary expansions. I know that the binary expansion of any real number is unique, however this does not seem to help. Would you have any suggestions?
 A: Suppose $x \in [0 , 1]$. Then either $x \in [0, \frac 1 2] = S^1_1$ or $x \in [\frac 1 2, 1] = S^2_1$. Now suppose $x \in S^k_n$ for some $k ,  n \in \Bbb N$. Then since $S^k_n = [\frac{k-1}{2^n},\frac{k}{2^n}] = [\frac{2k-2}{2^{n+1}},\frac{2k}{2^{n+1}}] = [\frac{2k-2}{2^{n+1}},\frac{2k - 1}{2^{n+1}}] \cup [\frac{2k - 1}{2^{n+1}},\frac{2k}{2^{n+1}}] $. Then it follows that $x \in S^{2k-1}_{n + 1}$ or $x \in S^{2k}_{n + 1}$. Notice $S^k_n \supset S^{2k-1}_{n + 1}$ and $S^k_n \supset S^{2k}_{n + 1} $ are proper subsets. By induction this proves that $x$ belongs to a series of nested intervals of the form $S^k_n$ and there are infinitely many.
As before $x \not \in  [0, \frac 1 2]$ or $x \not \in [\frac 1 2, 1]$. The set which does not contain $x$ contains infinitely many nested subsets of the form $S^k_n$ that also do not contain $x$. A formal construction as above may be required. 
A: You quite obviously have for every $n$ that $$
  \bigcup_{k=1}^{2^n} S_n^k = [0,1] \text{.}
$$
It follows immediately that if you fix $x$, then for every $n \in \mathbb{N}$ there's a $k_n$ with $x \in S_n^{k_n}$. Which yields (i), and even a particular sequence of $S_n^k$ which all include $x$, namely $S_1^{k_1}, S_2^{k_2},\ldots$
For proposition (ii), set $$
  \overline{S}_n^k = \begin{cases}
  S_n^{k+2} &\text{if $1 \leq k \leq 2^n - 2$} \\
  S_n^{1} &\text{if $k = 2^n - 1$} \\
  S_n^{2} &\text{if $k = 2^n$}
  \end{cases}
$$
and observe that $S_n^k \cap \overline{S}_n^k = \emptyset$ if $n \geq 3$. So if the $k_n$ are as above for some $x$, none of the following contain $x$ $$
  \overline{S}_3^{k_3},\overline{S}_4^{k_4},\ldots
$$
