So I think the proof is divided into three main parts, existence of the equation, remainder must be less than the divisor, and uniqueness of the quotient and the remainder. I will copy the proof here and cut in at the points where I have problem with understanding.
Definition:
Let a and b be integers with $b \gt 0.$ Then there exist unique integers q and r with the property that $a=bq+r$, where $0\le r\lt b$.
Proof:
We begin with the existence portion of the theorem. Consider the set
$S$ = {$a - bk$ | $k$ is an integer and $a-bk \ge 0$}.
If $0 \in S,$ then $b$ divides $a$ and we may obtain the desired result with $q=a/b$ and $r=0.$
Q1: Why is $q = a/b$? Isn't $q$ supposed to be an integer? His interchanging use of $k$ and $q$ makes me even more confused. Since both k and q are integers, shouldn't $q \neq a/b$ anyways?
Now assume $0\notin S.$ Since $S$ is nonempty [if $a\gt 0, a-b\cdot0 \in S;$ if $a \lt 0, a-b(2a)=a(1-2b) \in S; a\neq0$ since $0 \notin S$], we may apply the Well Ordering Principle to conclude that S has a smallest member, say $r=a-bq.$ Then $a=bq+r$ and $r\ge0,$ so all that remains to be proved is that $r\lt b$.
Q2: I don't understand how $0 \notin S$ implies $a \neq 0$. We could still have $a-bk \gt 0$ even if $a=0$ by choosing $k \lt 0.$
If $r\ge b$, then $a-b(q+1)=a-bq-b=r-b\ge 0,$ so that $a-b(q+1) \in S$. But $a-b(q+1) \lt a-bq$, and $a-bq$ is the smallest member of $S$. So, $r\lt b$.
To establish the uniqueness of q and r, let us suppose that there are integers $q, q', r,$ and $r'$ such that $$a=bq+r, \quad 0\le r\lt b, \quad and \quad a=bq'+r', \quad 0 \le r' \lt b.$$
For convenience, we may also suppose that $r' \ge r.$ Then $bq+r=bq'+r'$ and $b(q-q') = r'-r.$ So, $b$ divides $r'-r$ and $0 \le r'-r \le r' \lt b.$ It follows that $r'-r=0,$ and therefore $r'=r$ and $q=q'. \quad \blacksquare$
I understood the rest of the proof and would like some help with understanding the parts I remarked.
If $0\in S$, then there is some integer $k$ such that $a-bk=0$; call that integer $q$. Then $a-bq=0$, so $q=a/b$. This is the special case in which $a$ happens to be a multiple of $b$.
If $a=0$, then we can take $k=0$ to see that $0=a-b\cdot 0\in S$.