Statement:
Let $L_1, L_2, L_3, \ldots$ be a sequence of parallel lines in the plane, and put $X = \cup_{n\ge 1} L_n$. Define topology $\tau$ on $X$ by $G$ is $\tau$- open if $G = \varnothing $ or $\exists n_0 \in \mathbb N$ such that $L_n \setminus G$ is finite for every $n \ge n_0$.
Given an arbitrary sequence $(z_n)$ in $X$, verify either some subsequence of $(z_n)$ lies on one single line $L_q$ or there is a subsequence and a non-empty open set $G$ such that no term in the subsequence belongs to $G$.
Solution:
Looks like $\tau$ is cofinite meaning if $(L_n)$ is convergent, it's eventually constant. Suppose $(L_n)$ converges. Then there’s some natural $n_0$ s.t. $L_n = L_q$ for all $n \ge n_0$. That means every $L_{n_i} \in \{L_{n_0}, L_{n_1}, L_{n_2}, \ldots\}$ lies on $L_q$.
If $(L_n)$ does not converge to $L_q$, then there’s some neighborhood $G$ of $L_q$ s.t. $L_n$ are not in $G$ for any $n$.
Does this solution make sense?
edit:
Since $\tau$ is cofinite, convergent $(z_n)$ implies it's eventually constant. Suppose $(z_n)$ converges. Then there’s some natural $n_0$ s.t. $z_n = L_q$ for all $n \ge n_0$. That means every $z_{n_i} \in \{z_{n_0}, z_{n_1}, z_{n_2}, \ldots\}$ lies on $L_q$.
If $(z_n)$ does not converge to $L_q$, then there’s some neighborhood $G$ of $L_q$ s.t. $z_n$ are not in $G$ for any $n$.
The notation $(z_n)$ is quite confusing: the $n$ in $z_n$ has nothing to do with the $n$ in the lines $L_n$. I change it to $(z_k)$.
This almost has nothing to do with topology, just elementary set theory: For each $n$, let $Z_n = \{ k \in \mathbb N: z_k \in L_n\}$.
If $Z_n$ is infinite for some $n$, there is a subsequence $(z_{k_j})$ in the line $L_n$.
If not, then $Z_n$ is finite for all $n\in Z$. Let $C_n = \{ z_k : k \in Z_n\}$. Then $C_n$ is a finite set in $L_n$. Let $G = X\setminus \cup_n C_n$. Then there is no $z_k$ in $G$.