Let $a\mid bc $ then prove or disprove $a\mid (a,b)c$

I'm really confused about your solution. $a\mid bc$ doesn't imply that either $a\mid b$ or $a\mid c$ unless $a$ is a prime number. Anyway, the statement you try to prove/disprove is true. You can write $(a,b)=ka+lb$ when $k,l\in\mathbb{Z}$. Then $(a,b)c=kac+lbc$. $a$ divides $a$ and so $a\mid $. Also $a\mid bc$ which implies $a\mid lbc$. And hence $a$ divides the sum $kac+lbc=(a,b)c$.

The main problem in your proof is that knowing that $a|bc$ does not imply that $a|b$ or $a|c$.

Hint: Use the fact that the $\gcd(a,b)$ can be written as follows: $$ \gcd(a,b)=ax+by $$ for some integers $x$ and $y$. By substituting this into the expression $\gcd(a,b)c$, you get $$ \gcd(a,b)c=acx+bcy. $$ Both of the terms $acx$ and $bcy$ are divisible by $a$ (but for different reasons). Can you work from here?