$m\mathbb{Z}= \bigsqcup_{s=0}^{n-1} ms + mn\mathbb{Z}$ 
Claim: $m\mathbb{Z}= \bigsqcup_{s=0}^{n-1} ms + mn\mathbb{Z}$

May I know if my proof is correct? Thank you.
If $m=0,$ the claim is obviously true.
Now, let $m \neq 0.$
Suppose $\exists s,s'(\neq s) \in \{0,1,...,n-1\}$ such that $(ms + mn\mathbb{Z}) \bigcap (ms' + mn\mathbb{Z}) \neq \emptyset.$ 
$x \in (ms + mn\mathbb{Z}) \bigcap (ms' + mn\mathbb{Z}).$
$ \implies ms + mnr = ms' +mnr',$for some $r,r' \in \mathbb{Z}.$
$\implies n| s-s'.$ (Contradiction) 
Given $z\in m\mathbb{Z},$ there exists $y\in \mathbb{Z}$ such that $z=my.$ Also, $\exists q,r\in \mathbb{Z},$ with $0\leq r \leq n-1,$ such that $z= m(nq+r).$ Hence $z \in \bigsqcup_{s=0}^{n-1} ms + mn\mathbb{Z}.$ It is clear that $\bigsqcup_{s=0}^{n-1} ms + mn\mathbb{Z} \subseteq m\mathbb{Z}$
 A: The proof is correct, under the unstated assumption that $n>0$.
Here's a refinement, because contradiction is not necessary in the first part.
The case $m=0$ is obvious, so we can assume $m\ne0$. Suppose $0\le s<n$ and $0\le s'<n$ and that $z\in(ms+mn\mathbb{Z})\cap(ms'+mn\mathbb{Z})$. We can assume $s\ge s'$ without loss of generality; in particular $s-s'<n$. Then
$$
z=ms+mnr=ms'+mnr'
$$
for some $r,r'\in\mathbb{Z}$. Therefore $n\mid(s-s')$ with an obvious simplification, so $s=s'$. Thus the sets
$$
ms+mn\mathbb{Z}\quad(s=0,1,\dots,n-1)
$$
are pairwise disjoint. It is also clear that their union is a subset of $m\mathbb{Z}$.
If $mr\in\mathbb{Z}$, then $r=nq+s$, for some $q$ and $s$, with $0\le s<n$, so $mr=ms+mnq\in (ms+mn\mathbb{Z})$. Therefore $m\mathbb{Z}$ is contained in the union of those sets.

A different proof. It is well known that $\mathbb{Z}=\bigsqcup_{s=0}^{n-1}(s+n\mathbb{Z})$, because this is the partition relative to the congruence modulo $n$. Since the map $f\colon r\mapsto mr$ from $\mathbb{Z}$ to $m\mathbb{Z}$ is bijective (for $m\ne0$), we immediately get that
$$
m\mathbb{Z}=f(\mathbb{Z})=\bigsqcup_{s=0}^{n-1}f(s+n\mathbb{Z})
=\bigsqcup_{s=0}^{n-1}(ms+mn\mathbb{Z})
$$
