Proving the so-called "Well Ordering Principle"
You should be more detailed in what you claim. Your proof is wrong, and it is very hard to understand why you obtain said conclusions. My point is:
"Then, of necessity..." Why?
"...if $B$ contained even one $b∈\Bbb N$, then that $b$ would satisfy $b=\min(B)$? Why?
A correct proof would go as follows:
P Let $B\subseteq \Bbb N$ be nonempty. We prove by induction that $B$ has a least element. Assume by contradiction that $B$ has no least element. Let $J$ be the set of elements that are not in $B$. Since $0$ is a lower bound of $\Bbb N$, $0\notin B$ (else it would be a least element) so $0\in J$. We prove by induction that $0,1,\dots,n\in J\implies n+1\in J$. Indeed, suppose that $0,1,\dots,n\in J\implies n+1\in J$. Then $n+1$ cannot be in $B$ since then it would be a lower bound of $B$, and since $0,1,\dots,n\notin B$, it would be a least element. It follows $n+1\notin J$. By induction, $J=\Bbb N$ so $B=\varnothing$ which is impossible.
As Pete is saying, WOP is equivalent to PMI.
PROP Suppose every nonempty subset of $\Bbb N$ has a least element. Let $B$ be a subset of $\Bbb N$ with the following properties
$(1)$ $0\in B$.
$(2)$ $n\in B\implies n+1\in B$
We show that $B=\Bbb N$.
P Let $B$ be as above. Consider the set of $\Bbb N\setminus B$, and assume by contradiction it is not empty. By the WOP, it has a least element, call it $a$. Since $0\in B$, this element must be of the form $a=n+1$ for some $n\in \Bbb N$. Since $n+1$ is the first element that is not in $B$, $n$ is an element of $B$. But then $n+1\in B$, which is absurd. It follows that $\Bbb N\setminus B$ must be empty, so $B=\Bbb N$, as claimed.
What do you mean by "if $B$ contained even one $b \in \mathbb{N}$?" Your argument works if $B$ contains exactly one $b$. It doesn't seem to work otherwise: e.g. the subset $\{1,3\}$ contains "even one element $3$", but $3$ is not the least element.
Also, yes: you can take WOP as part of an axiomatic description of the natural numbers $\mathbb{N}$. You should know that -- for instance, given the other four Peano axioms -- it is equivalent to the Principle of Mathematical Induction.
Assume not. Then, of necessity, we'd have to have $B=\varnothing$, for if $B$ contained even one $b\in\mathbb{N}$, then that $b$ would satisfy $b=\min(B). $
This only works if $B$ contains exactly one element. For any other $B$, you have to assume that $\min$ is well-defined, i.e. every subset of $\mathbb{N}$ has a least member, so your reasoning is circular.