How would I prove that zero is the only element in the intersection of the following family of sets? Find the intersection of the following family:
$$
\mathcal{M}=\{n\mathbb{Z}:n\in\mathbb{N}\},
$$
where
$$
n\mathbb{Z}=\{\dots,-3n,-2n,-n,0,n,2n,3n,\dots\}
$$
for each $n\in\mathbb{N}$.
It appears that the only element that is common to all sets $M_n\in\mathscr{M}$ is $0$. Therefore, $\bigcap\limits_{n\in\mathbb{N}}M_n=\{0\}$. But how do I prove this? It occurred to me that since $M_n$ is a proper subset of the set $M_i$ for all $i\in\mathbb{N}$ such that $i<n$, the number of elements in the intersection will decrease as the index $n$ increases. Examining a finite number of cases, zero appears to be common to every set and the number of nonzero elements in the intersection seems to approach zero, but this is an inductive argument at best. I'm not sure how I would show $\bigcap\limits_{n\in\mathbb{N}}M_n=\{0\}$ deductively.

You already mentioned that $0 \in n \mathbb{Z}$ for all $n \in \mathbb{N}$. So we only need to show that for $k \neq 0$ there is some $n \in \mathbb{N}$ such that $k \not \in n \mathbb{Z}$. For this we can simply take $n = |k|+1$ (check this!). So indeed $\bigcap_{n \in \mathbb{N}} n \mathbb{Z} = \{0\}$.
Note that if you consider $0$ to be a natural number (i.e. $0 \in \mathbb{N}$) then the entire question trivialises, because $0 \mathbb{Z} = \{0\}$.
