My Analysis lecturer defined $\mathbb{N}$ as the smallest set such that $0 \in \mathbb{N}$ and if $n \in \mathbb{N}$ then $n+1 \in \mathbb{N}$. He then proceeded to "prove" the Principle of Mathematical Induction as follows: Let $P(n)$ be a family of propositions indexed on $\mathbb{N}$. Suppose that $P(0)$ is true and that for any $n \geq 0$, if $P(n)$ is true then $P(n+1)$ is true. Let $S$ be the set of natural numbers such that $P(n)$ is true. Then $S$ is a subset of $N$. Since $P(0)$ is true, $0 \in S$, and if $n \in S$ then $P(n)$ is true, so $P(n+1)$ is true, or equivalently $n+1 \in S$. Thus $S$ has the properties $0 \in S$ and if $n \in S$ then $n+1 \in S$; as $N$ is the smallest set with these properties, $N$ is a subset of $S$, but since we already know $S$ is a subset of $N$ it follows that $S=N$, i.e. $P(n)$ is true for all natural numbers.
My question is whether this proof is actually valid. I can't see any obvious flaw in the logic, but at the same time, I always thought that either the Principle of Mathematical Induction or (equivalently) the Well-Ordering Principle had to be taken as an axiom and couldn't be proved itself.
