If $a\mid b$ and $b\mid c$ then $a\mid c$ (for $b\neq 0$)
Given $(a\mid b) \iff (\exists{n} \in \Bbb{Z})\ a \cdot{n} = b$ and $(b\mid c) \iff (\exists{m} \in \Bbb{Z})\ b \cdot{m} = c$,
then $b = a \cdot n$ so substituting for $b$ we have $(a \cdot n) \cdot m = c $
which is equivalent to $a \cdot mn = c$. Therefore we have $(a\mid c)$.
On
It is correct and you can even note that the assumption $b\ne0$ is redundant.
In a different format.
Since $a\mid b$, there exists $m\in\mathbb{Z}$ with $b=am$; since $b\mid c$, there exists $n\in\mathbb{Z}$ with $c=bn$. Therefore $$ c=bn=(am)n=a(mn) $$ and we can conclude that $a\mid c$.
What about $b=0$? Stating $a\mid 0$ is not a problem, because it is true for every $a\in\mathbb{Z}$, since $0=a0$.
Stating $0\mid c$, on the other hand, says that $c=0$, because $c=0n$ implies $c=0$. The statement $0\mid 0$ is perfectly correct and there is no division by $0$ involved.
As you see, $a\mid0$ and $0\mid c$ implies $c=0$, and therefore $a\mid c$. But there is no point in making the special case in the general proof.
Your logic is sound, and your proof is fairly good, but you can polish up the argument:
Given $(a\mid b) \iff (\exists{n} \in \Bbb{Z})( a \cdot{n} = b),$ and $(b\mid c) \iff (\exists{m} \in \Bbb{Z})( b \cdot{m} = c),$
then $b = a \cdot{n}.$ And by substituting $a \cdot n$ for $b$, we have $(a \cdot n) \cdot m = c,$
which is equivalent to $a \cdot(mn) = c.$
Since $m, n \in \mathbb Z$, so too is $mn\in \mathbb Z$.
Therefore, we have $(a\mid c)$