Prove that if $\gcd (a, b) = 1$ then for any $d$ such that $d \mid a$ we have
$$\gcd (d, b) = 1$$
We're given $\gcd (a,b) = 1$. By Bezout's Identity this means there exist integers $x,y$ such that
$$ ax + by = 1 $$
We're also given $d \mid a$, which means there exists an integer $z$ such that
$$ a = zd $$
Putting these together, we have
$$ (zd)x + by = 1 $$
More explicitly,
$$ d(zx) + by = 1 $$
Here, the $\gcd(d,b)$ must also divide 1, and that is only possible if $\gcd(d,b)=1$.