I know this is statement is false and only correct for prime numbers. However, before I found the obvious counter examples I wrote a proof by contradiction and I cannot find the error I made. My proof is as follows:
a, b, c are integers and a is not $0$
Negation: $a | bc$ and $(a \nmid b$ and $a \nmid c)$
From the left side I know $bc\mod a = 0$
From the right side I can tell $b\mod a \neq 0$ and $c \mod a \neq 0$ so $bc \mod a \neq 0$ because if $a$ doesn't divide $b$ or $c$, then it won't divide any multiple of $b$ or $c$.
Now I have a contradiction. I have not been able to find my error, although the original statement is obviously false.
The issue is in your final step. If $b$ and $c$ don't divide $a$, it is still possible for their product to divide $a$! In fact, this is a restatement of the very theorem you are trying to prove (written in terms of modular arithmetic instead of division).
As the comments have mentioned, $4 \not \mid 2$ and $4 \not \mid 6$, but $4 \mid 12$. This is the same as saying $2$ and $6$ are not $0$ mod $4$, yet their product $12$ is $0$ mod $4$.
Hope this helps ^_^