= Solution
An action assigns each $g\in G$ a <homeomorphism> of $X$, with $e\cdot x=x$ and $g\cdot(h\cdot x)=(gh)\cdot x$. For the covering assertion use a properly discontinuous free action: every $x$ has an open neighbourhood $U$ such that $gU\cap U=\varnothing$ for every $g\ne e$. Then the translates $gU$ are pairwise disjoint. The <quotient map> $p$ is open because $p^{-1}(p(U))=\bigcup_g gU$ is open. Each restriction $p:gU\to p(U)$ is a continuous open bijection and hence a <homeomorphism>. Thus $p(U)$ is evenly covered and \b[$p:X\to X/G$ is a <covering map>].
For the surjective <homomorphism> one additionally needs $X$ path connected; choose $x_0$ and its image $y_0$. Lift a based loop at $y_0$ from $x_0$. Its endpoint has a unique form $gx_0$. Homotopy lifting makes this assignment well-defined on the <fundamental group>. With concatenation meaning first the first loop and then the second, the second lift from $gx_0$ is $g$ times the lift from $x_0$, so the endpoint is $ghx_0$. This proves the <homomorphism> property. For each $g$, a path in $X$ from $x_0$ to $gx_0$ projects to a loop realizing $g$. Hence
$$
\boxed{\pi_1(X/G,y_0)\twoheadrightarrow G.}
$$
Pointwise freeness alone is insufficient for a covering: an irrational rotation action of $\mathbb Z$ on the circle is free but its orbits are dense. Path connectedness is also essential for the final surjectivity claim as written: the discrete space $X=G$ with left translation is a free properly discontinuous action whose quotient is a point, and its trivial <fundamental group> cannot surject onto nontrivial $G$. The proof supplies the standard connected covering-space interpretation rather than hiding these hypotheses.
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