If $a\mid b$, then $\forall x(b\in \mathbb z) \exists y(a \in \mathbb Z/a\mathbb Z([x]b\subseteq [y]a.$

Question

I've got a question on how to prove the following statement:

If $a\mid b$, then $\forall \big([x]_b\in \mathbb Z, \exists ([y]_a \in \mathbb Z,([x]_b\subseteq [y]_a)\big).$

If additionally, $a\neq b$, then "$\subseteq$" in the statement above can be substituted with "$\subsetneq$".

So for example let $a = 3,\; b = 6, \;x = 1,\; y = 1$.

1. $3\mid 6$
2. $[1]_6 = [7,13,19,25,31,37, ...]$ and $[1]_3 = [4,7,10,13,16,19, ...]$
3. So $[1]_6 \subseteq [1]_3$ seems to be true but how would one prove that?

It makes totally sense but unfortunately I don't know how to prove it. Any tips/ideas?