I want to show that this is true. Here's my approach.
Let $(a,b)=d$ be the greatest common divisor of $a$ and $b$. Since $a|b,$ there exists an integer $k$ such that $b=ak$ Thus $(a,ak)=d$.
Since $(a,ak)=d$, there exist integers $s,t$ such that $$as+akt=d$$ $$a(1s+kt)=d$$ $$a(1,k)=d$$ $$a\cdot 1=d$$ $$a=d$$
Is this a correct proof?
No. There's no reason to believe that $1s+kt = \gcd(1,k)$, because it's not true for pretty much any $s,k,t$ you could pick.
The best way to approach this problem is probably straight from the definition of $\gcd$. You know that $a \mid a$ and $a \mid b$, and for any $d$ such that $d \mid a$ and $d \mid b$, we have that $d \mid a$; so $a$ is the greatest common divisor of $a$ and $b$.
$s + kt = 1$ is not necessarily true. Let $d = (a,b)$, then:
$d | a$. But $a | a$ and $a | b$, so $a | (a,b) = d$. So
$a = d$
More directly, you must have $d \mid a$. Also, $a \mid a$ and $a \mid k a$, so $a$ is a common divisor, and so $a \le d$. Hence $a = \pm d$.
You can probably do it in a way similar to yours, but I think an easier way would be to note that $a$ is a divisor of both $a$ and $b$, so it is a common divisor of $a$ and $b$. To show that $a = \gcd(a, b)$, suppose for sake of contradiction that $\gcd(a, b) = d$ such that $d > a$.
If this were true, can $d$ actually be a divisor of $a$?
