What you will demonstrate in this exercise is that there is a very natural equivalence between well-ordering and induction.
You did a good job by defining $C$. But, in fact, $C$ is empty, which is what you need to prove. So, by means of contradiction, you need to assume that it is not. Then, as you suggest, it has a minimal member $k$. Now, you have to split into two cases: either $k=1$ or $k-1$ is a positive integer that is not in $C$ The first case would be the base case of an induction proof, and the second is essentially the induction hypothesis that will lead to a contradiction when you conclude that $k\notin C$ by doing the algebra.
Well, this is a proof where you should just use what you know - you recognize that the well-ordering principle tells you something about a least element, so let's investigate that! There's nothing special about a largest element* (especially since $a_0 \geq a_1 \geq a_2 \ldots$ implies that $a_0$ is the largest... which wasn't very interesting).
So, let's suppose we have a sequence $a_0\geq a_1\geq a_2 \geq \ldots$. To apply the well-ordering principle, we need to come up with a non-empty set - and the set $\{a_0,a_1,a_2,\ldots\}$ is the only thing in sight, so let's use the well-ordering theorem on that! This means that there is some $a_n$ such that $a_n \leq a_m$ for all $m$ - and now we're basically done! Look at any $m\geq n$. We know (or can prove by induction) that since $a_n \geq a_m$ since the sequence is non-increasing. However, we just used this magic principle to choose $n$ so that $a_n \leq a_m$. That means that $a_n=a_m$! In particular, following this line of reasoning gives us the following:
Given any non-increasing sequence of natural numbers, there is some term $a_n$ after which the sequence is constant.
Which says that, no matter how you start, you'll end with something like $c,c,c,c,c,c,c,c,\ldots$ where the term $c$ goes on forever.
*Saying there's nothing special about a largest element is somewhat of a lie. You can prove the statement I highlighted by induction on $a_0$ - if $a_0=0$, we are done because there are no smaller natural numbers. Then, you can note that this implies that if the sequence ever hits $0$, it must be constant - but then if you investigate $a_0=1$, you find that either the sequence eventually hits $0$, or it repeats $1$ forever - and you can twist this into a formal inductive argument. This is not the easiest path, however. You can also do things like suppose that $a_0>a_1>\ldots$ is strictly decreasing and show that it can have at most $1+a_0$ terms, or just look at the finite set $\{0,1,2,\ldots,a_0\}$ and ask how many times each term repeats.
Best Answer
When you're given a set, you cannot just state it is non-empty by picking an element from the set. In fact, to be able to say $x \in N,$ you need to justify why you can do that.
Many times, it is trivial to show that the set is non-empty. There also are times, when the given assertion trivially holds even for the empty set you can split off the proof into cases when the given set is empty or non-empty. In the latter case, you can pick an element from the set.
So for example, you have two sets $A$ and $B$ and the assertions says $A\subseteq B.$ Then you take two cases.
Case 1: $A = \emptyset.$ Then $A \subseteq B$ is trivial.
Case 2: $A \neq \emptyset.$ Then you pick $x \in A$ and use any known information to conclude $x \in B.$
You cannot say "Let $x \in A$," since $A$ maybe the empty set.