the line with two origins and the related equivalence relation

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I'm trying to prove that the line with two origin is not Hausdorff but I'm having trouble understanding the problem statement. $\textbf{I'm not asking for a solution}$, I just need a clarification of what the problem means. Here is the problem:

(Problem 3-16 from LeeTM): Let $X$ be the subset $(\mathbb{R} \times \{0\}) \cup (\mathbb{R} \times \{1\}) \subseteq \mathbb{R}^2$. Define an equivalence relation on $X$ by declaring $(x,0) \sim (x,1)$ if $x \neq 0$. Show that the quotient space $X/ \sim$ is locally Euclidean and second countable, but not Hausdorff.

where LeeTM:=Introduction to Topological Manifolds by John Lee.

$\textbf{My question:}$ what does "declaring $(x,0) \sim (x,1)$ if $x \neq 0$" mean? I have two guesses.

(Guess 1) Let $(x_1,y_1),(x_2,y_2) \in X$, we say $(x_1,y_1) \sim (x_2,y_2)$ if and only if $x_1,x_2$ are non-zero with $x_1=x_2$ and $\{y_1,y_2\}=\{0,1\}$

But if this is what he meant, then the equivalence class of $(0,0) \in X$ is the empty set because its $x$-coordinate is zero : no point in $X$ can have the same non-zero $x$-coordinate as $(0,0)$. However, equivalence classes can't be the empty set because of the reflexivity. Therefore, I had a second guess

(Guess 2) $(x_1,y_1) \sim (x_2,y_2)$ if and only if $x_1=x_2$ and $\{y_1,y_2\}=\{0,1\}$.

In this case, we have

$$(0,0) \notin [(0,0)] = \{(0,1)\} $$ which contradicts the reflexivity of equivalence relation.

So my second guess is also wrong. $\textbf{What does he mean really? what is the equivalence relation he's trying to define?}$

In example 3.48 of LeeTM, he used the similar wording: " let $\sim$ be the equivalence relation on $\bar{\mathbb{B}}^2$ ( the closed unit disk in $\mathbb{R}^2$) generated by $(x,y) \sim (x,-y)$ for all $(x,y) \in \partial \mathbb{B}^2$". He defined the equivalence relation for the points on the boundary but what about those on the interior? (Likewise, in problem 3-16, he defined the equivalence relation for $x \neq 0$ but what about $x=0$?)

He uses this "description" more than one time so I think it may be a terminology that everyone knows (except me). $\textbf{Can someone explain to me what does this wording/description of equivalence relation mean?}$

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They mean the equivalence relation generated by $(x, 0) \sim (x, 1)$ for all $x \neq 0$. That is: $(x_1, y_1) \sim (x_2, y_2)$ when either (1) $x_1 = x_2$ and $x_1 \neq 0$; or (2) $x_1 = x_2$ and $y_1 = y_2$ (the points are the same).

This is also known as the smallest equivalence relation on $X$ in which $(x, 0) \sim (x, 1)$ for all $x \neq 0$. Reflexivity means that each point must be equivalent to itself, and in addition we impose the given extra identifications.

Edited to add: intuitively, in topology, we usually think of this as a "gluing" or "collapsing". For example, suppose you take the closed unit interval $[0, 1]$ and then take the quotient space by the equivalence relation defined by $0 \sim 1$. That glues the two ends of the interval together, which topologically gives you a circle.