Let $g= \gcd (a, b)$ and $d\mid a$, $d \mid b$. Prove that $d \mid g$.
The premises $d\mid a$ and $d \mid b$ mean that for some integers $x,y$ we have
$$ a = x d $$
$$ b = y d $$
Substituting
$$ g = \gcd(xd, yd) $$
That is, for some integers $m,n$
$$ g =mxd $$
$$ g =nyd $$
This gives us $d \mid g$.