If $\frac{m}{n}$ cannot be written in lowest terms, this means that for all $m', n'$ such that $\frac{m'}{n'} = \frac{m}{n}$, $m'$ and $n'$ share some common divisor $d > 1$.
The later statement says that if
$$\frac{\frac{m_0}{p}}{\frac{n_0}{p}}$$
can be expressed in lowest terms, as some $\frac{q}{r}$, then as
$$\frac{q}{r} = \frac{\frac{m_0}{p}}{\frac{n_0}{p}},$$
it is also true that
$$\frac{q}{r} = \frac{m_0}{n_0},$$
and so $\frac{q}{r}$ would be an expression of $\frac{m_0}{n_0}$ in lowest terms. Since no such expression can exist, since $m_0 \in C$ and by definition of $C$, there can also be no way to express
$$\frac{\frac{m_0}{p}}{\frac{n_0}{p}}$$
in lowest terms.
Call the positive integer $k$ bad if there is a set of $k$ envelopes, numbered $0$ to $k-1$, with contents as described in the question, and an integer $n\lt 2^k$, such that no subset of the envelopes contains exactly $n$ dollars between them.
We want to prove there are no bad $k$. Suppose to the contrary that there are. Then the set $B$ of bad $k$ is non-empty. It follows that there is a smallest bad $k$. Call it $m$.
$m$ is greater than $0$, because $m=0$ holds obviously.
So $m$ is bad, and all positive integers $\lt m$ are good.
We arrive at a contradiction by showing that $m$ is good, that is, that any number $n\lt 2^m$ of dollars can be obtained by choosing an appropriate collection of the envelopes.
Case (i), $n\lt 2^{m-1}$: Since $m-1$ is good, there is a subset of the first $m-1$ envelopes that gives sum $n$.
Case (ii), $2^{m-1}\le n\lt 2^m$. Take the contents of the biggest envelope, that is, envelope with label $m-1$, which has $2^{m-1}$ dollars.
Then we still need an amount $a$ of dollars, where $0=2^{m-1}-2^{m-1}\le a\lt 2^m-2^{m-1}=2^{m-1}$. Because $m-1$ is good, we can produce this amount $a$ from the first $m-1$ envelopes. Hence by adding the contents of the envelope with label $m-1$, we get our $n$ dollars.
Remark: This is just an ordinary proof by induction, rewritten to use "well-ordering principle" language.
Best Answer
If $x^2=2y^2$, then $(2y-x)^2=2(x-y)^2$.