Suppose that $X$ is the set of all lines $$L_{n} = \mathbb R \times \{n\} \quad\textrm{ for $n\in\mathbb Z^+$}$$ and $Z$ is the set of all lines that cross the center of plane and its slope is positive: $$(L_n)' = \{(x,nx) :\, n \in \mathbb Z^+\}.$$ Show that $g:X \to Z$ defined by $g((x,n)) = ((x,nx))$ it is not a quotient map.
By definition, If I show that for some subset of $Z$ like $W$, $p^{-1}(W)$ is open in $X$, but $W$ is not open in $Z$, then it is not quotient. I don' know how to show it.
Let us do it for closed $p^{-1}(W)$. For $n \in \mathbb Z_+$ let $R_n = \{ x \mid \lvert x \rvert \ge 1/n^2 \} \subset \mathbb R$, $M_n = R_n \times \{n\} \subset L_n$ and $M'_n = \{ (x,nx) \mid x \in R_n \} \subset L'_n$. Note that $p$ maps $M'_n$ bijectively onto $M'_n$ so that we have $p^{-1}(M'_n) = M_n$.
Now let $W = \bigcup_{n \in \mathbb Z_+} M'_n$. This set is not closed because $(0,0) \notin W$, but $(0,0) \in \overline{W}$. Note that $(1/n^2,1/n) \in M'_n \subset W$. We have $p^{-1}(W) = p^{-1}(\bigcup_{n \in \mathbb Z_+} M'_n) = \bigcup_{n \in \mathbb Z_+} p^{-1}(M'_n) = \bigcup_{n \in \mathbb Z_+} M_n$. This set is closed, thus $p$ is not a quotient map.