Sequence on $\mathbb{Z}$ with the evenly spaced integer topology I find some problems in solving this exercise:

Consider $\mathbb{Z}$ with the evenly spaced integer topology. Prove that the sequence $a_n = n!$ converges to zero.

My reasoning is: let $S(a,k)=a + k \mathbb{Z}= \left \{ a+kn: n \in \mathbb{Z} \right \}$, with $a,k \in \mathbb{Z}$ be a generic arithmetic progression. Then, we have to prove that for any sequence $S(0,k)$ we have $a_n \in S(0,k)$ definitively.
But, how can we prove the last part, i.e., $\exists n_0$ such that $a_n \in S(0,k) \quad \forall n >n_0$? And how can we show a counterexample, for instance $a_n$ does not converge to 3?
 A: Because $n\geqslant k\implies n!\in S(0,k)$, since $k\mid n!$ then.
And you don't have $\lim_{n\to\infty}n!=3$ (or any other number different from $0$), because $\Bbb Z$, endowed with that topology is separated, and therefore a sequence can have, at most, one limit.
A: Your first question has already been answered, so I will address the second.
To show $a_n$ does not converge to $3$, it suffices to find an arithmetic progression $S(3,k)$ such that for every $n_0$, there exists $n>n_0$ such that $a_n\notin S(3,k)$. In other words, we can find terms arbitrarily far out in the sequence $a_n$ that are not in $S(3,k)$.
Take $k=2$ so $S(3,k) = \{...,-5,-3,-1,1,3,5,...\}$. Then we can find arbitrarily large terms in the sequence $a_n=n!$ that are not in $S(3,k)$ since for all $n \geq 2$, $n!$ is even.
This exact argument won't work for every possible choice of limit (only odd limits). I'd have to put in a bit more thought to come up with a direct proof that $a_n$ does not converge to, say, $2$ or $4$. However, the integers with the evenly spaced topology are a Hausdorff space, and so limits of sequences must be unique.
