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Define an relation $\sim$ on the lines of a plane $\mathcal{U}$ as follows: $l\sim m$ if and only if either $l=m$ or $l$ and $m$ are parallel (no intersection).

Prove that $\sim$ is an equivalence relation when $\mathcal{U}=\mathbb{R}^2$.

Let $l,m,n$ be lines

First the relation is reflexive since $l=l$ thus $l \sim l$. Now Assume $l \sim m$ thus either $l=m$ or $l$ and $m$ are parallel. Thus in both cases symmetry holds since if $l=m$ then $m=l$ and if $l$ and $m$ are parallel then $m$ and $l$ are parallel thus $m \sim l$. Finally if $l \sim m$ and $m \sim n$ thus $l=m$ and $m=n$ or $l,m,n$ are all parallel to each other. Now in the first case if $l=m$ and $m=n$ then $l=n$ and $l \sim n$. If $l$ and $m$ are parallel to each other and $m$ and $n$ are parallel to each other it follows that $l$ is parallel to $n$ therefore again $l \sim n$ so transitivity holds. Therefore the relation is an equivalence relation.

Now I also want to consider why the equivalence relation fails for a hyperbolic plane model.

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    What do you define as parallel in hyperbolic geometry? – 2017-02-01
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    It's because the parallel postulate fails. You can have more than one line through a given point, both parallel to a given line (but neither equal to nor parallel to each other). – 2017-02-01

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It's easy to see in the half-plane model for the hyperbolic plane.

Let $l_1$ and $l_2$ be two distinct vertical lines (perpendicular to the $x$-axis), and let $c1$ be a half-circle (with origin in the $x$-axis) that intersects $l_1$ but not $l_2$. Then $l_1$ and $l_2$ are parallel, $c$ and $l_2$ are parallel, but $c$ and $l_1$ are not (they intersect).