Proof:

Let $\tilde{x} \in {\tilde X}$ be arbitrary and $x = p({\tilde x})$. Choose an evenly covered neighborhood $U$ of $x$ and $\tilde{U}$ be a sheet lying over $U$ and containing ${\tilde{x}}$. Then $\tilde{U}$ is an open set in ${\tilde X}$ containing ${\tilde x}$ that is mapped by $p$ homeomorphically onto $U$. Thus $p$ is a local homeomorphism and we have proved (i). Let ${\tilde G}$ be an arbitrary open set in ${\tilde X}$. Then ${\tilde G}$ can be covered by open sets ${\tilde U}$ such that $p$ maps each ${\tilde U}$ homeomorphically onto an evenly covered open subset $U$ of $X$ (why?). Then $G = p({\tilde G})$ is the union of such evenly covered neighborhoods $U$ implying that $G$ is an open set in $X$. Thus $p$ is an open mapping. Finally to prove (iii) suppose that $\tilde{z} \in {\tilde X}$ is a limit point of $p^{-1}(x)$. Pick an arbitrary evenly covered neighborhood $U$ of $z = p({\tilde z})$ and a sheet ${\tilde U}$ lying over $U$ containing ${\tilde{z}}$. In particular the restriction of $p$ to the sheet ${\tilde U}$ is injective. But since ${\tilde z}$ is a limit point of $p^{-1}(x)$, this sheet must contain infinitely many points of $p^{-1}(x)$ which means $p$ restricted to ${\tilde U}$ cannot be injective which is a contradiction.