Let $\mathscr{G}=\prod_\chi S^1$, where the product is indexed by every character of $\mathbb{Z}$ (which is a morphism $\mathbb{Z}\to S^1$). If $j:\mathbb{Z}\to\mathscr{G}$ is defined by $n\mapsto (\chi(n))$, then $G=j(\mathbb{Z})$ is algebraically isomorphic to $\mathbb{Z}$ and inherits a structure of topological group from $\mathscr{G}$.
In Nondiscrete topology making $(Z,+)$ a topological group. we proved that $G$ is not discrete.
Even though $G$ is not discrete, it is true that in $G$ every convergent sequence is eventually constant. I know how to prove it using measure theory but I think there may exist an easier proof of it and would like to know if any of you knows how to do it.
(Using measure theory we can prove that for every sequence of integers $(a_n)$ that is not eventually zero, there is a character $\chi$ such that $\chi(a_n)$ does not converge to $1$. Our result then follows.)
I'm very confused. It seems your question is very easily equivalent to: show that if $(a_n)_n$ is a sequence of integers that is not eventually $0$, then there is a character $\chi$ of $\mathbb{Z}$ such that $(\chi(a_n))_n$ does not converge to $1$. You have a measure theory proof of this fact and want a non-measure theory proof of it. All the stuff not in parentheses seems like unnecessary fluff.
Here's an easy proof (without measure theory). Obviously if $(a_n)_n$ takes a nonzero value infinitely often, we're done. So we may suppose $(a_n)_n$ goes off to infinity. By sparsifying, we may suppose $a_n > 2^{2^n}$ for each $n$, say. Then there is some $\theta \not = 0$ such that $||a_n\theta||_{\mathbb{R}/\mathbb{Z}} \to \frac{1}{2}$ as $n \to \infty$ (exercise). Taking $\chi$ to be $n \mapsto e(n\theta)$ then finishes the job, as $\chi(a_n) \to -1$ as $n \to \infty$.