This is related to Bézout identity.
Let $c_0=as_0+bt_0=\min(S)\tag{0}$
Then if $d$ is a common divisor of $a,b$ then $a=da'$ and $b=db'$ we have $as_0+bt_0=d(a's_0+b't_0)=c_0$
$\text{d divides a,b}\Rightarrow d\mid c_0\tag{1}$
On the other hand for $c\in S$, we have $c=as+bt=c_0q+r=(as_0+bt_0)q+r$
So $r=(s-s_0q)a+(t-t_0q)b\ \overset{?}{\in} S\quad$ but $0\le r<c_0$ by euclidian division definition.
If $r>0$ this contradict the fact that $c_0$ is $\min(S)$, so $r=0$ and $c=c_0q$
Note: since $q\ge 1$ then $c\ge c_0$ ($q\ge 0$ integer, and cannot be $0$, else $c=0\notin S$). We already know it from the minimum assertion, but it's good to have it back as a verification.
$c\in S\Rightarrow c\text{ is a multiple of }c_0\tag{2}$
Using the same method of euclidean division by $c_0$, we can show that $a$ and $b$ are also multiples of $c_0$. (note: $q,r$ are dummy variables, they are not the same than previously).
$a=c_0q+r=(as_0+bt_0)q+r$ with $0\le r<c_0$ so $r=(1-s_0q)a+(-t_0q)b\overset{?}{\in} S$ and we conclude like previously, same with $b$.
$a,b\ \text{ are multiples of } c_0\tag{3}$
Combining $(0),(1),(2),(3)$ we proved that $c_0=\gcd(a,b)$ and that $c_0$ is the smallest number reachable by a relation of type $S$ whose elements are all multiples of $c_0$.
To put this in everyday words: if two numbers are multiple of $3$ for instance, then their sum and their difference is also a multiple of $3$. Plus if they are not equal, the smallest possible difference is $3$.
Replace $3$ by $\gcd(a,b)$ and this is exactly what is mathematically written above.
$P(m)$: For any nonnegative integer less than $2^{m+1}$, there is a selection of
$\quad\quad\;$ envelopes whose constants add up to exactly that number of dollars.
Let $S = \{m : m \in N$ and $P(m)$ is false $\}$
If $S$ is not empty, by W.O.P, $S$ has a least element $m_0$
Since $m_0$ is the least element of $S$, $m_0 - 1$ is not in $S$
So, $P(m_0 - 1)$ is true and there is a selection from envelopes $0, 1, ..., (m_0 - 1)$ that sum to $t$ for $0 \le t < 2^{m_0}$
Now, suppose we have envelopes $0, 1, ..., (m_0 - 1), m_0$
By selecting envelope $m_0$ and then selecting from envelopes $0, 1, ..., (m_0 - 1)$ there is a selection that sums to $t$ for $2^{m_0} + 0 \le t < 2^{m_0} + 2^{m_0}$ or $2^{m_0} \le t < 2^{m_0+1}$
This means that with $m_0$ envelopes, there is a selection that sums to $t$ for $0\le t < 2^{m_0+1}$
This means $P(m_0)$ is true which is a contradiction that $m_0$ is the least element of $S$.
So, $S$ must be empty and $P(m)$ must be true for all $m \in N$
Best Answer
Let me begin by saying that we do not assume that the statement is true for all values $n$ such that $c < n$. I think you may have misspoken in writing that.
Your last sentence is correct. We assume that the statement is true for all $n < c$, and we are required to deduce from this that it is also true for $c$, whence a contradiction.
The reason we don't talk about the other values $n > c$ for which the statement is true, is that we know little about them. Our only assumption is that $c$ is the smallest value for which the statement is false. This does allow us to conclude that the statement is true for values $n < c$, but usually allows us to say nothing about values $n > c$.
In specific instances, it may be possible to deduce something about values $n > c$ from the assumption that $c$ is the smallest number for which the statement is false, and any method of deriving a contradiction will do the job. But this is not the case in general.