You seem to have misplaced a quantifier. It is true, as you said, that $p(k)\implies p(k+1)$ is logically equivalent to $(\neg p(k))\lor(p(k)\land p(k+1))$, and therefore the second premise of the induction axiom, $(\forall k)\,[p(k)\implies p(k+1)]$, is equivalent to $(\forall k)\,[(\neg p(k))\lor(p(k)\land p(k+1))]$. But then you went from this statement to
$$
[(\forall k)\,\neg p(k)]\lor[(\forall k)\,[p(k)\land p(k+1)].
$$
This "distribution of $\forall$ over $\lor$ is not valid. The correct formula allows the possibility that $\neg p(k)$ holds for some values of $k$ and $p(k)\land p(k+1)$ holds for all the other values of $k$. The transformed formula requires the same alternative to hold for all $k$, and this is something quite different.
You don't have to show that one positive integer is false in order to show that the proposition is false.
Notice that the quoted proof says:
assume that there is at least one positive integer for which $P(n)$ is false.
That's something of a clue already. In mathematics, you can't prove something by first assuming that it's true. Instead, this is an example of a proof by contradiction. We start out by assuming that what we are trying to prove is false, and then derive a contradiction. Since we assume that the axioms of mathematics are sound, this means that our original assumption must be false, and hence that what we are trying to prove is true.
In this case, we want to prove that $P(n)$ is true for all $n$. The negation of that statement is that $P(n)$ is false for some $n$. We now work with this assumption and try to derive a contradiction.
We now let $m$ be the smallest number such that $P(m)$ is false. It's very important that $m$ is the smallest such example, which is why we need the well-ordering principle. In order to make the dependence on the well-ordering principle explicit, we say:
Let $S$ be the set of $n$ such that $P(n)$ is false. By assumption, $S$ is non-empty, so it must have a least element $m$ by the well-ordering principle.
But you don't have to think that way. Intuitively, if there is one counterexample among the natural numbers, there must be a smallest one. (Note that the same is not true if we restrict attention to the integers, which do not satisfy the well-ordering principle. For example, $\{-1, -2, -3, \dots\}$ hasn't got a least element.
Now consider $P(m-1)$. Since $m$ is the smallest number with $P(m)$ false, and $m-1$ is smaller than $m$, $P(m-1)$ must be true.
The reason you can't have both $P(4)$ and $P(5)$ false when $m$ is $5$ is that $m$ has to be the smallest counterexample, and it clearly isn't in this case, as $4$ is a smaller counterexample. In this case $m=4$ (or some smaller number).
$P(m-1)$ implies $P(m)$ by assumption. We are trying to show that $P(n)$ is true for all $n$ when we are given these pieces of information:
- $P(1)$ is true.
- If $P(k)$ is true then $P(k+1)$ is true.
We obviously need some information about $P$ to prove that $P(n)$ is true for all $n$, and this is the information we are given. Now, setting $k=m-1$, and noting that $P(m-1)$ is true, we must have that $P(k+1)=P((m-1)+1)=P(m)$ is true.
We have now arrived at a contradiction. This serves to disprove our initial assertion that $P(n)$ is not true for some $n$. It does not disprove the theorem that we are trying to prove - that $P(n)$ is true for all $n$. Indeed, since it disproves the negation of that theorem, it proves the theorem and we win.
Best Answer
Long comment
The Principle of Mathematical Induction is often an axiom because it is quite obvious :
It is obvious but ... the above is not a proof : the "goes on forever" is not what we "like" in mathematical proofs.
The same for the corresponding argument :
Conclusion : there are many "Principles" that we can use interchangeably, in the sense that, assuming one of them we can prove the others, but we have to assume some of them in order to rigorously license the "form of argument" typical of natural numbers that lies behind the "goes on forver".
This "goes on forver" is exactly what "identifies" natural numbers : start from a "first" one (zero or one) and then add one : we get a new number. Then add one... We can go on forever and we'll always find a new number.
This intuitive concept is rigorously charcterized by the so-called Peano axioms :