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 $x \in S$.
$S$ 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?
Hint: What can you say about the set $-S$?

Notice that, as others have pointed out, $0$ need not belong to $S$.