Well-Ordering Principle Let S be a nonempty subset of $\{n \in \mathbb{Z}: n \le 0\}$ Let $S$ be a nonempty subset of $\{n \in  \mathbb{Z}: n \le 0\}$. Use the Well-Ordering Principle to show that $S$ has a
greatest element; in other words, prove there exists $s \in  S$ such that $s \ge  x$ for all $x \in S$. 
I can clearly tell that the greatest element here is $0$ just by looking at the set. My problem comes by way of what this question is asking me to do. Am I supposed to simply identify $0$ as the greatest element? If so, here is my attempt:
let $S$ be a nonempty subset of  $\{n \in  \mathbb{Z}: n \le 0\}$, if $s \in  S$, then $s \in \{n \in  \mathbb{Z}: n \le 0\}$ The greatest element of $S$ is $0$. Thus $s \ge x$ for all $x \in S$. 
 A: $S$ can be any subset of the non-positive integers!
For example, if $S=\{-1,-2,-3 \dots\}$, then in fact $-1$ is the maximal element.
but there is still hope to use your idea.
Indeed, if $S$ were all of $- \mathbb N$, then $0$ would be the maximal element. But since it is not, assume for contradiction, that it has no  maximal element. Then $0 \notin S$, since if it were, then since $S \subset -\mathbb N$, all of its elements would be less than zero as well. Then, $S \subset - \mathbb N\setminus \{0\}$. In this case, what is the maximal element of $-\mathbb N\setminus \{0\}$ can $-1 \in S$ if $S$ has no maximal element? 
A: As user296602 pointed out in the comments, $0$ will not always be the greatest element. If your set consists of $\{-10, -5, -8 \}$ then your greatest element is $-5$.
The well-ordering principle says:

If $S$ is a subset of $\{n \ge 0|n \in \mathbb{Z} \}$, then there exists an element $M \in S$ such that $\forall s \in S$ we have $M \le s.$

Now somehow you have to use this principle to prove that a set $A$ of non-positive integers has a greatest element. But the well ordering principle deals with sets of non-negative integers, so we somehow have to find an interesting set which is related to our set but has only non-negative integers. 
Consider the set $B= \{-a : a \in A\}$. Now this set satisfies the hypothesis (the if part) of the well-ordering principle, so we can apply it to the set. So $B$ has an element $N$ such that $\forall b \in B$ we have $M \le b$. But every $b$ is equal to $-a$ for some $a \in A$. 
Can you take it from here?
A: Hint: What can you say about the set $-S$?
Notice that, as others have pointed out, $0$ need not belong to $S$.
