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! 

The following argument offers an alternative proof; the contrast between different proofs can be instructive.
For any $k$, define
$\tag 1 R_k = \{\sum_{n=1}^k \frac{x_n}{2^n} \; | \; x_n \in \{0,1\} \land \sum_{n=1}^k \frac{x_n}{2^n} \le x\}$.
This is a finite set with at most $2^k$ elements.
Proposition 1: For every $k$,
$\tag 2 x - \text{max}(R_k) \le 2^{-k} $
Proof
To get a contradiction, assume that $x_n$ defines the maximum number in $R_k$ and
$\tag 3   x - \sum_{n=1}^k \frac{x_n}{2^n} \gt 2^{-k} $
It can't be true that $x_n = 1$ for all $n$, since then $x \gt 1$. But then since
$\tag 4   \sum_{n=1}^k \frac{x_n}{2^n} + 2^{-k} \le x $
the $2^{-k}$ number can be 'absorbed' into the sigma expression form, so that a another number can be found in $R_k$ greater than the number defined with the $x_n$. $\quad \blacksquare$
It is now easy, using $\text{(2)}$, to show that the increasing sequence $\text{max}(R_k)$ converges to $x$.

Note: Although induction wasn't explicitly used, the $\sum$ notation (an 'induction machine') is carrying the proof construction. 
