While going through 6.042J from MITOCW, in the text Mathematics for Computer Science, I came across the following problem at which I'm stuck. 
Now, I proceeded doing the proof in the following manner. Suppose $C$ be the set of counterexamples to $P(n)$ where
$$P(n)::=\text{It is possible to make a postage stamp of n}¢$$ So, $$C = \bigg\{n \in \mathbb{N}, n \geq50 \mid \text{P(n) is false} \bigg\}$$ where $\mathbb{N}$ denotes the set of non-negative integers (i.e., integers $\geq 0$)
We assume that $C$ has a least element $m \in C \text{ such that P(m) doesn't hold}$.
Assuming that $\text{51¢, 52¢, 53¢, ..., 56¢}$ are all makeable (given in the question), we can say that $$m > 56$$ $$\implies m-6 > 50$$
Now, as $\text{m is the least element in C, P(m$-6$) has to be true}$
So, $(m-6)¢$ is makeable. Thus, adding $6¢$ to it, $(m-6)¢+6¢$ is also makeable, and thus $m¢$ is also makeable.
Thus we arrive at a contradiction, i.e., $\text{P(m) is actually true}$. So, $C$ must be empty. Thus it is possible to make any amount of postage over $50¢$.
Is this proof complete and correct? Why did they say to assume that all denominations from $51¢ \text{ to } 100¢ $ were makeable, when in fact, assuming upto $56¢$ would suffice?
Also, for part $\text{(b)}$, isn't $49¢$ actually makeable? Because $49¢ = 14¢+14¢+21¢$ So why does it say to prove that it is not? Is this part incorrect?
One way of explaining the problem composers intent, on the first part of the problem is that
Since $50$ is makeable, and since the range $51$ through $100$ is makeable, you have an immediate proof by induction.
That is, assuming that the range $(50k + 1)$ through $(50k + 50)$ is makeable, and that $(50)$ is also makeable, you immediately conclude that the range $(50[k+1] + 1)$ through $(50[k+1] + 50)$ is also makeable.
So, the problem composer is trying to stretch the intuition of the Math student into visualizing the induction principle, applied in as simple a way as possible.
As you indicated, you can conclude that since $51$ through $56$ is makeable, and $6$ is makeable, then an alternative approach is feasible. The problem composer is attempting to coerce you into following the specific induction path, that he has laid out.
As you indicated, $(49)$ is makeable, so this is a clear mistake on the part of the problem composer. As it turns out, each of $44$ through $48$ is also makeable. So, the problem composer, having overreached, should have settled for asking you to show that $43$ is not makeable.
Let $a,b,c$ denote the non-negative integer scalars to be applied to $6, 14,$ and $21$, respectively.
If $c = 2$, clearly $(43- [2 \times 21])$ will not be makeable.
Suppose that $c = 1$.
Then, you have to find scalars $a,b$ so that $6a + 14b = (43 - 21) = 22.$ Neither $b = 0$, nor $b = 1$ will work in this situation.
Suppose that $c = 0$.
Then, you have to find scalars $a,b$ so that $6a + 14b = (43).$ This is impossible because $(43)$ is odd, while $6$ and $14$ are even.
Therefore, there is no satisfying combination that has $c \in \{0,1,2\}$
Therefore, $(43)$ is not makeable.