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! 

All that is necessary is to highlight/expand on what the OP has already written out.
Let $P(n)$ be the inductive hypothesis and we write as true
$\tag 1 P(n): \quad 0 \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}, \; \text{ with } x_1,...x_n \in \{0,1\}$
We need to set $x_{n+1} = 0$ or $x_{n+1} = 1$  such that $P(n+1)$ is true, i.e.
$\tag 2 0 \le x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$
Case 1: $x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$
In this case $\text{(2)}$ will be true by simply setting $x_{n+1} = 0$.
Case 2: $x - \sum_{k=1}^n \frac{x_k}{2^k} \ge \frac{1}{2^{n+1}}$
Using $\text{(1)}$ we can write
$\tag 3  \frac{1}{2^{n+1}} \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}$
Subtracting $\frac{1}{2^{n+1}}$ we can write
$\tag 4  \frac{1}{2^{n+1}} - \frac{1}{2^{n+1}} \le [x - \sum_{k=1}^n \frac{x_k}{2^k}] -\frac{1}{2^{n+1}} < \frac{1}{2^n} - \frac{1}{2^{n+1}}$
or 
$\tag 5  0 \le [x - \sum_{k=1}^n \frac{x_k}{2^k}] -\frac{1}{2^{n+1}} < \frac{1}{2^{n+1}}$
So if we set $x_{n+1} = 1$ statement $\text{(2)}$ will be true.
