Let 𝑥 =𝑝(˜𝑥) =𝐺˜𝑥 ∈𝑋. Since 𝐺 acts properly discontinuously, ˜𝑥 has an open neighbourhood ˜𝑈 such 𝛾1˜𝑈 ∩𝛾2˜𝑈 =∅ for any 𝛾1,𝛾2 ∈𝐺 with 𝛾1 ≠𝛾2. Then 𝑈 =𝑝(˜𝑈) =𝐺˜𝑈 is an evenly covered open neighbourhood of 𝑥: 𝑝−1(𝑈) =𝑝−1(𝑝(˜𝑈)) = ⨿𝛾∈𝐺𝛾˜𝑈 and 𝑝 ↾𝛾˜𝑈 :𝛾˜𝑈 →𝑈 is a homeomorphism for each 𝛾 ∈𝐺, since it is surjective and continuous (by construction), injective (no to points in ˜𝑈 lie in the same orbit by proper continuity), and open (in fact 𝑝 is open, since ˜𝑉 open implies 𝛾˜𝑉 open for all 𝛾, so ⋃𝛾𝛾˜𝑉 =𝑝−1𝑉 is open and thus 𝑉 =𝑝(˜𝑉) is open). Thus every 𝑥 ∈𝑋 has an evenly covered neighbourhood, so 𝑝 is a covering of the orbit space 𝑋 with ˜𝑋 itself.