Sets by Inclusion Clarification \begin{align}
A &= \lbrace 70, 210, 280\rbrace\\
B &= \mathbb{Z}\\
C &= \lbrace n\in\mathbb{N}\,|\, n=7 m \,\mbox{for some} \, m\in\mathbb{Q} \rbrace\\
D &= \lbrace n\in\mathbb{N}\,|\, n=35 m \,\mbox{for some} \, m\in\mathbb{N} \rbrace \\
E &= \lbrace n\in\mathbb{N}\,|\, n=7 m \,\mbox{for some} \, m\in\mathbb{N} \rbrace
\end{align}
$$ Letter ⊆ Letter⊆Letter ⊆Letter ⊆Letter $$
I'm starting discrete mathematics and encountered this problem where we are tasked with ordering the set by inclusion. I understand the premise, but I think that my reasoning is wrong. Upon initial glance, I would sort it as $ A⊆ D⊆E⊆B⊆C$, as A only has the three elements, the elements of D can be made by E, but then things start to get confusing with B and C. B stipulates all integers, which I understand, but C would be able to "create" elements that are not integers by virtue of the m $\in \Bbb{Q}$ allowing the use of all rational numbers, as well as all of the integers that could exist in $\Bbb{N}$. However, would this even matter as we are adding them to set $n\in\Bbb{N}$, which is only natural numbers? Any help would be appreciated. I've also included a picture for clarification.
A screenshot of the problem

$A$ by inspection and $C,D,E$ by definition are subsets of $\Bbb N$, and $\Bbb Z$ contains all of $\Bbb N$.
$C$ is in fact just equal to $\Bbb N$ as $n \in \Bbb N$ means $n = 7\cdot \frac{n}{7}$, so $7$ times a rational.
$A$ consists of multiples of $35$, so $A \subseteq D$.
A multiple of $35$ is always a multiple of $7$ of course, so $D \subseteq E$.
So in all we get $$A \subseteq D  \subseteq E  \subseteq C  \subseteq B$$
