About the proof that every real number in the unit interval is the limit of a sequence of dyadic numbers 
Given $x \in (0,1)$, show there exists a sequence $(x_n) \subset \{0,1\}$ such that $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$.

After running into difficult in trying to solve this problem, I found this MSE post, which, to my chagrin, gave me more difficulties. I'm trying to follow John Ma's reasoning. Here's how I understand it: 

If $x < \frac{1}{2}$, the choose $x_1 =0$ and therefore $|x-\frac{x_1}{2}| = |x| = x < \frac{1}{2}$, and in fact $x-\frac{x_1}{2} \ge 0$, so in this case we can find $x_1$ for which $0 \le x - \frac{x_1}{2} < \frac{1}{2}$. If, however, $x \ge \frac{1}{2}$, then choose $x_2 =1$, and $|x-\frac{x_1}{2}| < \frac{1}{2}$ holds if and only if $- \frac{1}{2} < x - \frac{1}{2} < \frac{1}{2}$ if and only if $0 < x < 1$. Since $0 < x < 1$ is in fact true, so must $|x-\frac{x_1}{2}| < \frac{1}{2}$. Also, we have $x - \frac{x_1}{2} \ge 0$ since we assumed $x \ge \frac{1}{2}$. Hence, in either case we found $x_1 \in \{0,1\}$ for which $0 \le x-\frac{x_1}{2} < \frac{1}{2}$ holds. Now assume that $x_1,...x_n \in \{0,1\}$ have been chosen such that $0 \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}$. From this we want to show that it is possible to choose $x_{n+1} \in \{0,1\}$. If $x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$, then choose $x_{n+1} = 0$ and we obtain $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$. If, however, $x - \sum_{k=1}^n \frac{x_k}{2^k} \ge \frac{1}{2^{n+1}}$, choose $x_{n+1} = 1$. Then $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} = (x - \sum_{k=1}^n \frac{x_k}{2^k}) - \frac{1}{2^{n+1}}$

But here's the problem: why is $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k}$ nonnegative? If that can't be shown, then I cannot appeal to Squeeze lemma to show $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$. I could really use some help. Thanks in advanced! 

(1).Let $x_1=0$ if $0<x<1/2$ and $x_1=1$ if $1/2\leq x<1.$ Let $y_1=x_1/2.$ We define the sequences $(x_n)_{n\in \Bbb N}$ and $(y_n)_{n\in \Bbb N}$ as follows:
(2). Suppose $y_n\leq x.$ Let $x_{n+1}=0$ if $x<y_n+2^{-(n+1)}.$ Let $x_{n+1}=1$ if $x\ge y_n+2^{-(n+1)}.$ And let $y_{n+1}=y_n+x_n2^{-(n+1)}.$ In both cases we have $y_n\leq x\implies y_{n+1}\leq x.$
And we have $y_1\leq x.$ So by induction we have $y_n\leq x$ for all $n$.
(3). We have $y_n\geq x- 2^{-n}\implies y_{n+1}\geq x-2^{-(n+1)}.$ Proof: Suppose $y_n\geq x-2^{-n}.$ Then 
$\quad$(i).If $y_n\leq x-2^{-(n+1)}$ then $x_{n+1}=1$ so $\quad y_{n+1}=y_n+2^{-(n+1}\geq (x-2^{-n})+2^{-(n+1)}=x-2^{-(n+1)}.$ 
$\quad$(ii). If $y_n>x-2^{-(n+1)}$ then $x_{n+1}=0$ so  $y_{n+1}=y_n>x-2^{-(n+1)}.$
In both (i) and (ii) we have  $y_{n+1}\geq x-2^{-(n+1)}.$
And we have $y_1\geq x-2^{-1}.$ So $y_n\geq x-2^{-n}$ for all $n$ by induction.
(4). Since $x-2^{-n}\leq y_n\leq x$ for all $n,$ we have $x=\lim_{n\to \infty}y_n=\sum_{n=1}^{\infty}x_n2^{-n}.$
Remark. In (2) we use recursion and induction together, to define $x_{n+1}$ and $y_{n+1}$ recursively from $y_n$, with the inductive hypothesis that $y_n\leq x.$  If you wanted, you could instead say " If $y_n>x$ then let  $x_{n+1}=1=y_{n+1}$ " ( or some other arbitrary values) for a purely recursive def'n, and then prove separately that $y_n\leq x$ for all $n$ by induction.
