Let $\Gamma$ be discrete in S. Then for any region $Ω$ in $S$, $Ω \cap \Gamma$ is discrete in Ω

I'm trying to understand this existence of triangulation's proof in this book. I have problems to understand the lemma 8.2.6:

Let $\Gamma$ be discrete in S. Then for any region $Ω$ in $S$, $Ω \cap \Gamma$ is discrete in Ω

I didn't understand why this $\gamma$ in the proof exists.

I need help here.

Thanks

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The set $\Gamma\cap\Omega$ is a different set of curves than $\Gamma$. One curve in $\Gamma$ could become many curves in $\Gamma\cap\Omega$. This is how we find $\gamma$.
More specifically, we have a point $x$ with a neighborhood $U$ (guaranteed by discreteness of $\Gamma$ in $S$) that meets only finitely many curves in $\Gamma$, but (by assumption toward contradiction) meets infinitely many curves in $\Gamma\cap\Omega$. Since each curve in $\Gamma\cap\Omega$ comes from some curve in $\Gamma$, there must be a curve in $\Gamma$ which induces infinitely many curves in $\Gamma\cap\Omega$. This curve is $\gamma$.
Look at $\gamma^{-1}\Omega$ and $\gamma^{-1}U$ in $\mathbb{R}$. Each should have countably many connected components. Pick a point from each. – Neal Nov 15 '12 at 7:08
what do you mean by $\gamma^{-1}\Omega$ $\gamma^{-1}U$? – user42912 Nov 15 '12 at 17:16
Since $\gamma:\mathbb{R}\to S$, we can look at the inverse images of $U$ and $\Omega$ under $\gamma$. – Neal Nov 15 '12 at 17:45