I tried this: we have to prove $\gcd(b,c)=1$. $a$ is prime to $b$, then $\gcd(a,b)=1$. And at least one of $a$ or $b$ is prime. $c$ is a divisor of $a$ then $\gcd(c,a)=c$. And $a=kc$ where $k$ is an integer. It proves that $a$ is not prime. So $b$ is prime. Now since $b$ is prime, then $\gcd(b,c)=1$.
Is it correct?
Let's $\gcd(c,b) = k$ then $k|b$ and $k|c$. Since $c|a$ we have $k|a$, so $k$ is a common divisor for $a$ and $b$. But since $a$ and $b$ are coprime, $k \le 1$ and so $k=1$.
An alternative proof: You can take advantage of the Bezout property of the integers. That is gcd$(a,b)=1$ if and only if there exist integers $s,t$ with $as+bt=1$.
If gcd$(a,b)=1$, then there exists $s,t \in \mathbb{Z}$ with $as+bt=1$. If $c \mid a$, then $a=cd$. But then we have $(cd)s+bt=1$ which implies $c(ds)+bt=1$, so gcd$(c,b)=1$.
