Consider the following theorem:
"Every non-empty set of positive integers has a minimum element".
The proof I usually see is one that uses contradiction, and does not seem like the easiest possible proof. I think there is an easier proof, and I wonder why I never see it. Does it contain an invalid assumption? The proof goes as follows:
First, prove by induction that the theorem is true for finite sets. Base case: It's true for a set of size 1, trivially. Induction step: Consider a finite set $S$ of size $n+1$. Let $s$ be a member of $S$. By I.H., $|S-s|$ has a minimum $s'$. If $s\lt s'$, then $s$ is the minimum of $S$. Otherwise, it is $s'$.
Second, let $T$ be a any non-empty set of positive integers ($T$ could be infinite). Let $t$ be an element of $T$. Consider the set $T\cap [0,t]$. The set is clearly finite, so it has a minimum element $min$. Next, we show that $min$ is also a minimum element of $T$. Let $x$ be an element of $T$. If $t\lt x$, then $min\leq x$. Otherwise $x\in T\cap [0,t]$, so $min\leq x$.
Is there a problem with this proof? I think when people prove that N is well-ordered, they do it in set theory books at a point where very little has been proven, so they can't assume much. Am I assuming too much in this proof? If not, why don't people ever use this very simple proof?