Proving if $d_0$ is the smallest positive integer in $S$ then $d_0 = \gcd(a,b)$

a. I'm not quite sure how you can just conclude from $r\in S$ that $d_0|s$ holds. But you can easily do it. Due to the euclidean division you get either $r=0$, which would imply $d_0|s$, or $0<r<d_0$, which is a contradiction to the minimality of $d_0$.

b. Use the last statement and $a,b\in S$.

c. Because $d_0\in S$ there exist integers $x,y$ with $d_0=ax+by$. So if $d|a,b$, then also $d|ax+by=d_0.$

d. We already showed $d_0|a,b$, which impliess $d_0|gcd(a,b)$, by the definition of the gcd. But $gcd(a,b)|a,b$, which impliess using c. that $gcd(a,b)|d_0$. So we have two positive numbers and each divides the other one, meaning they are equal, i.e. $d_0=gcd(a,b).$