My textbook states the Well-Ordering property as following:
If $A$ is any nonempty subset of $\mathbb{Z}^+$, there is some element $m \in A$ such that $m \le a$, for all $a$ in $A$ ($m$ is called the minimal element).
Okay, here is an outline of my proof:
Begin with the set $S$ which is a non-empty subset of $\mathbb{Z}^+$, and contains the two arbitrary elements $m$ and $a$. We can now compare these integers. There are three possible comparisons, and only one can be satisfied: either $m < a$, $m > a$, or $m = a$.
Suppose $m=a$, then clearly we can assert that $m \le a$, and thus the element $m$ is less than or equal to every element in the set $S$, and $S$ contains a minimal element. In fact, if we write $m \le a$ as $m \le m$, we can easily perceive that single element $m$ is the minimal element, in the case when $S$ contains a single element
Now, suppose that that the two elements in the set $S$ compare in the way such that $m <a$. Again, we can assert that $m \le a$, and clearly $m$ is the least element of $S$, because $m$ is less than or equal to every element in $S$.
The case in which $m >a$ is rather simple, so I will leave it out. The following is what I believe to be the most critical part of the proof (although, the entire proof itself may be flawed):
Suppose the add another element to the set $S$, call this arbitrary element $c$. We can use the same sort of comparisons as we did in the two when $S$ contained two elements, by implementing the transitive property, and arrive at the same conclusions as we did in the previous case.
We could then argue that, if we were to continue this process of adding elements to $S$, we could construct the arbitrary set $A$, which was alluded to in the statement of the Well-Ordering property, and thus show that it has some minimal element.
Does this seem to be a correct use of reasoning?
NOTE: This problem comes from Dummit and Foote's Algebra text. After I came up with this proof, I realized that the problem explicitly asked me to use induction. But I wanted to see if my proof my correct.
It's not clear what $S$ is supposed to be.
Moreover you didn't use the fact that you are working with natural numbers, you only used that the order is total (i.e. for every $a,b$ holds $a\leq b$ either $b\leq a$). So, if your proof were correct, you would have proved that every totally ordered set has the Well-Ordering property, that is obviously false.
You should use induction to prove this property: I give you a hint.
Start with a set $A \subseteq \mathbb{Z}_+$ and suppose it has no minimum. Then $0 \notin A$, otherwise it would be the minimum. But if $0 \notin A$ then $1 \notin A$, otherwise it would be the minimum. Etcetera...
So $A$ must be the empty set.