We were given the following theorem and proof in a lecture. I don't quite understand portions of it and would like some help.
Theorem. For each $n \in \mathbb Z$, let $F_n$ be a closed subset of $(n,n+1)$. Then $F= \bigcup_{n\in \mathbb Z}F_n$ is closed. Proof. Any convergent sequence from $F$ must eventually lie in the interval $(n-1,n+1)$ so all but finitely many terms lie in $F_{n-1}\cup F_n$ for some $n$. Since $F_{n+1} \cup F_n$ is closed, the limit must be in one of the two and so the limit is back in $F$. Thus, $F$ is closed
The following line is the one I don't understand completely.
Any convergent sequence from $F$ must eventually lie in the interval $(n-1,n+1)$
How was the specific interval obtained? What property of convergence sequence is my lecture using? Are they using "Cauchyness"?
If $(x_m)_{m \in \mathbb N}$ converges, let $l$ be its limit. Take some integer $n$ such that $l\in(n-1,n+1)$ and take some $\varepsilon > 0$ such that $(l-\varepsilon,l+\varepsilon)\subset(n-1,n+1)$. Then, if $m$ is large enough, $$x_m\in(l-\varepsilon,l+\varepsilon)\subset(n-1,n+1).$$