Proof for $\gcd$ being the smallest linear combination of $a,b \in \mathbb {Z}$. Please vet my approach:
First need prove that for the least positive value combination (let, $l$), it will divide $a, b \in \mathbb{Z}$. 
Will use the indirect, or contradiction approach to prove that $l \mid a$.  Similar will extend to $l \mid b$.
So, for $l \nmid a, \exists q, r \in \mathbb {Z}$, s.t. $a = al +r, 0 \lt r \lt l$.
Hence, $r = a - q.l = a - q(ax_0 + by_0) = a(1 - qx_0) - qby_0$. So, $r$ is a linear combination of $a, b$ with integer multipliers, let $x,y,$ where $x = 1 - qx_0, y = -qy_0$
The assumed form for $l = ax_0 + by_0$. Given, by contradiction that $0 \lt r \lt l$, have the following inequalities, as the integer multipliers for $r$ must be smaller than of $l$, and positive too:
=> $1- qx_0 \lt x_0$ - (i)  & $-q \lt 1$ - (ii)
=>  $1 \lt x_0(1+q)$ - (i)  & $q \gt -1$ - (ii)
Substituting for $q$ from (ii) into (i), get:
$ 1 \lt x_0($negative value$)$ -- obviously false, hence proved that $l  \mid a$ by contradiction.

Addendum In view of responses recd., have modified as follows. First, assume that $a,b \gt 0$. Second, form four cases of possible values of the integer multiplier, and will try to prove by contradiction in each.
Case (i) : $x_0 \gt 0, y_0 \gt 0$ : Have already proved this case above.
Case (ii) : $x_0 \gt 0, y_0 \lt 0$ : Implies $-qy_0 \lt y_0$, but as $y_0 \lt 0$, so cancelling $y_0$ from both sides leads to: $-q \gt 1 => q \lt -1$.
The two inequalities are :  $1 \lt x_0(1+q)$ - (i), $q \lt -1$ - (ii)
Substituting for $q$ from (ii) into (i), get: $1 \lt x_0$(negative value),  again obviously false.
Case (iii) : $x_0 \lt 0, y_0 \gt 0$ : Implies the two inequalities are: 
$1- qx_0 \lt x_0$ - (i), $q \gt -1$ - (ii)
$1 \lt x_0(1+q)$, but as $x_0 \lt 0$, so it changes to $-1 \gt x_0(1+q)$, for positive $x_0$. Now, this is also not possible as $(1+q)\gt 0$.
Case (iv) : $x_0 \lt 0, y_0 \lt 0$ : As assume $a, b \gt 0$, so this case is not possible.
Hence, proved for all possible values of $x_0, y_0$.
 A: Here is what I understood from your question: you want to show that if $l = \min\{ ax+by > 0 : x,y \in \mathbb{Z} \}$ then $l \mid a$ and $l \mid b$. Assuming $ l \nmid a$, you write $a = ql + r$ where $ 0 < r < l$, and thereby arrive at $r = a(1-qx_0) + b(-qy_0)$, where $l = ax_0 + by_0$.
Now, you argue that since $0 < r < l$, the integers appearing in the linear combination for $r$ must both be positive, and they must be smaller than the respective integers appearing in the linear combination for $l$. This is incorrect. It appears that you are assuming that $x_0$ and $y_0$ are both positive, but this is simply not true. Try computing the linear combinations for positive integers $a$ and $b$, and you will see that $x_0$ and $y_0$ often have opposite sign. It also appears that you are assuming that if one linear combination is smaller than another linear combination, then the same can be said for the respective coefficients. Again, this is not true in general, even if the coefficients are all positive. For example, let $a=1$ and $b=2$. Then $5a+b < 4a+2b$. You can find many more examples yourself by playing around with numbers.
Instead, here is what you need to do to complete the proof. Since $0 < r < l$, and $r$ is expressed as an integral linear combination of $a$ and $b$, it implies that $l$ is not the smallest positive integer that can be expressed as a linear combination of $a$ and $b$. This is a contradiction. Hence, our original hypothesis that $l \nmid a$ is false.

Regarding your modified proof, note that you are still making the assumption that if $ax+by < ax’ + by’$, then $x < x’$ and $y < y’$. As I mentioned in the second paragraph above, this is not true in general, regardless of the signs of $x$, $y$, $x’$ and $y’$. So, none of the four cases actually prove the desired result. In fact, since nothing can be said in general about the relative sizes of $x$ and $x’$, and $y$ and $y’$, from the fact that $ax + by < ax’ + by’$, therefore your approach cannot lead to a solution, by my understanding.
The simpler way to conclude the contradiction is as I mentioned in my third paragraph, above.
