Is the Groupoid of germs associated to an orbifold a Hausdorff proper Lie groupoid?

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I was studying the book Introduction to Foliations and Lie Groupoids by I. Moerdijk and J. Mrcun and I have a doubt. On page 140 they give the following definition for a proper Lie groupoid.

A Lie groupoid $\mathcal{G}$ is said to be proper if it is Hausdorff (that means the space of arrows is Hausdorff) and the map $(s,t): \mathcal{G}_1 \rightarrow \mathcal{G}_0 × \mathcal{G}_0$ is proper.

On page 141 they give the following definition and proposition

Let $Q$ be an orbifold and $U = {(U_i , G_i , φ_i)}_{i\in I}$ an orbifold atlas of $Q$. Put $X = \cup_{i\in I} U_i$ and $φ =\cup {φ_i }: X → Q$. Now let $Ψ(U)$ be the pseudogroup on $X$ of all transitions $f$ on $X$ for which $φ \circ f = φ|dom(f )$. Define the effective groupoid $Γ(U)$ associated to the orbifold atlas $U$ to be the effective groupoid associated to the pseudogroup $Ψ(U)$ (that means the groupoid of germs of transition functions), $Γ(U) = Γ(Ψ(U))$.

Proposition 5.29 Let $U$ be an orbifold atlas of an orbifold $Q$. Then

$(1)$ $Γ(U)$ is a proper effective groupoid.

$(2)$ If $U$ is an orbifold atlas of an orbifold $Q$ and $U$ is an orbifold atlas of an orbifold $Q$ , then $Γ(U)$ and $Γ(U )$ are weakly equivalent if and only if $Q$ and $Q$ are isomorphic.

My question is about item $(1)$ of the proposition. By definition to verify that $\Gamma(U)$ is proper we need to prove that the space of arrows of $\Gamma(U)$ is Hausdorff and that the map $(s,t): \mathcal{G}_1 \rightarrow \mathcal{G}_0 × \mathcal{G}_0$ is proper. But, if I am not mistaken, in the book it is proved that the second condition holds, but they tell us nothing about Hausdorffness of $\Gamma(U)$. And, to that matter since $\Gamma(U)$ is a groupoid of germs, I think it is rarely Hausdorff.

I would like to know wether $\Gamma(U)$ is Hausdorff or not, and if it is not, are they just relaxing on their definition of properness?

Any comment or help is appreciated. If you want me to clarify something please ask me to.