Proof:

Pick distinct points $\bar{y}, \bar{z}\in Y/G$ and let $U$ and $V$ be disjoint neighborhoods of $y$ and $z$ such that for every pair of distinct elements $g^{\prime}, g^{\prime\prime} \in G$,

$$g^{\prime}U \cap g^{\prime\prime}U = \emptyset, \quad g^{\prime}V \cap g^{\prime\prime}V = \emptyset.$$

Then

$$\eta^{-1}(\eta(U)) = \bigcup_{g\in G} gU,\quad \eta^{-1}(\eta(V)) = \bigcup_{g\in G} gV.$$

Since $G$ is a group of homeomorphisms, it follows from the definition of quotient topology that $\eta(U)$ and $\eta(V)$ are open sets containing $\bar{y}$ and $\bar{z}$. It is easy to see that $\eta(U)$ and $\eta(V)$ are disjoint and (i) follows and also that $\eta$ is an open mapping. Now $\eta$ restricted to each $gU$ is a continuous, open bijection, that is a homeomorphism onto $\eta(U)$ and so (ii) follows. Conclusion (iv) follows from (iii). To prove (iii) first observe that the map

$$\phi_g: y \mapsto g\cdot y, \quad y \in Y$$

is a deck transformation for each $g\in G$. The map

$$\psi : g \mapsto \phi_g$$

is easily seen to be a group homomorphism. To see that it is surjective, let $\phi$ be a deck transformation and $y_1$ be a given point in $Y$ and $\phi(y_1) = y_2$. Since $y_1$ and $y_2$ are in the same fiber, there is a unique element $g$ of the group such that $g\cdot y_1 = y_2$. Then the deck transformations $\phi$ and $\phi_g$ agree at $y_1$ and so are identical which means $\psi(g) = \phi$ proving surjectivity. If $g\in\;$ker$\;\psi$ then

$$\phi_g(y) = g\cdot y = y, \quad \forall y \in Y.$$

Since the action is properly discontinuous (and hence fixed point free) this forces $g = 1$.