As part of the course's assignments, we received a task to prove the following sentence using only Bézout identity:
Every common divisor of $a, b$ divides the gcd $(a, b)$.
I tried the following proof: By Bézout's identity, we know that $$\text{gcd}(a,b) = ax + by\tag{*}$$ for some integers $x,y$.
Let $c$ be the common divisor of $a$ and $b$. By definition, since $c$ divides $a$, we know that there exists $k_1$ so that $ck_1 = a$. The same is true for $b$, so $k_2$ exists so that $ck_2 = b$. Replace $a$ and $b$ in an equation $(*)$ and received:
(Edited) $$\text{gcd}(a,b) = cxk_1+cyk_2 = c(xk_1+yk_2)$$ Therefore we obtained that $c$, which is any common divisor of $a$ and $b$ divides gcd $(a, b)$.
Is proof enough? To my mind, it seems too simple. I'm pretty new in the elementary number theory world, I'd be happy to have another opinion. Thanks