How to prove that (k,r) is unique? The proof is about proving that there is a unique couple $(k,r)$ that $a = k*b + r$ and that $r < b$.
Let us now show that this pair $(k, r)$ is unique. Suppose instead that there exists $(k', r') \neq (k, r)$ such that $a = k '* b + r ' = k * b + r$, with $r' <b$ and $r <b$. As a consequence $(k-k') - b = r'- r$.
1) if $k = k'$ then $r'-r = 0$. So, $r'= r$ and $(k', r) = (k, r)$ for a contradiction. 
2) if $k > k'$ then $b \leq (k'-k) * b <r'-r \leq r'$. So, $b \leq r'$ for a contradiction. 
3) if $k < k'$ then $b (k - 'k) * b = r-r' \leq r$. So, $b \leq r$ for a contradiction.

This is an extract of a proof. 
Given the extract how is $(k-k') - b = r' - r$ obtained?
Why in case 2) does $r' - r \leq r'$?
Why in case 3) does $r - r' \leq r$?
 A: Based on what you wrote, there are several typos & missing information in the proof, e.g., it's assuming the values are integers, with $b$ being a positive integer. This is trying to prove, by contradiction, the uniqueness part of Euclidean division which, as the Wikipedia article says

Its main property is that the quotient and remainder exist and are unique, under some conditions.

From
$$a = k' \times b + r' = k \times b + r \tag{1}\label{eq1}$$
Adding $-k \times b - r'$ to the middle & right sides, and then factoring the left side below, gives
$$k' \times b - k \times b = r - r' \implies (k' - k) \times b = r - r' \tag{2}\label{eq2}$$
Thus, the "$- b$" part in "$(k-k') - b = r' - r$" should be "$\times b$" instead.
The proof forgot to state that $0 \le r$ and $0 \le r'$, otherwise the rest of the proof doesn't make sense. With this, case (2)'s $r' - r \leq r'$ is because $r \ge 0$, so subtracting it from $r'$ makes the result either the same or less than $r'$. Similarly, since $r' \ge 0$, case (3)'s $r - r' \leq r$ also holds.
Note with case (2)'s "$b \leq (k'-k) * b <r'-r \leq r'$", the "$<$" should be "$=$" instead. Also, there's an extra factor of "$b$" in the case (3) statement in the "$b (k - k') * b$" part.
A: The way to think of it is this:
we have $a = bk + r$.  
Now let's make the $k$ just one bigger by replacing it with $k' = k+1$.  Then we get $a + b = bk' + r = b(k+1) + r$.  But that is too big.  
To get it back down to $a$ we have to subtract $b$.
$a = bk' + (r-b) = b(k+1) + (r-b)$.  But now that remainder is $r-b$ and because $r < b$ that is a negative number.  It is not true that  $0 \le r-b < b$.  That's just wrong.
So we can't do $a = b(k+1) + R$ where $0\le R < b$ because the only $R$ that will work is $R = r-b< 0$.
And if we can't replace $k$ with $k+1$ we certainly can't replace $k$ with $k + m$ for $m > 0$.
That'd give us if $a = b(k+m) + R$ then $R$ has to equal $r - bm$ and as $bm \ge b$ than would give us that $R=r-bm< 0$.
And if we try to replace $k$ with $k-1$ or $k -m$ we run into a similar problems.
$a = b(k-1) + R$ and $a=bk + r$ means that $R$ has to equal $r + b$.  But $r+b \ge b$.  And that's not acceptable.  And $a = b(k-m) + R$ would mean $R = r +bm > b$ and that isn't acceptable.
SO the $k$ is the only number we can have if $a=bk + R$ where $0\le R < b$.
And if $k$ is the only number we can have then $r = a-bk$ is the only $r$ we can have.
The proof is doing the exact same thing.  Only slicker and better.
====
See if you can follow:
If $a = bk + r$ and $a = bk' + r'$ then
$bk + r = bk' + r'$ so
$bk = bk' + (r'-r)$ so
$bk -bk' =r'-r$
$(k-k')b = r'-r$.
If $k > k'$ then $k-k' \ge 1$ and $(k-k')b \ge b$.
So $r'-r = (k-k')b \ge b$.
So $r' \ge b+r \ge b$ which contradicts that $r' < b$.
If $k < k'$ we get a similar problem
$(k-k')b = r'-r$ so 
$(k' -k)b =r-r'$ so $k'-k\ge 1$ and $r-r' = (k'-k)b \ge b$ so $r \ge b+r' \ge b$ which contradicts that $r' < b$.
So the only option is $k'= k$.  But that mean $r' = r$.
So $a = bk +r$ is theonly option if we require that $0 \le r < b$.
(Note: if we don't have that requirement there are infinitely many pairs but $r$ can be negative or bigger than $b$ or ...)
