$\mathbb{R}P^n$ is an $n$-manifold: how to show locally Euclidean and Hausdorff properties? 
Show that the real projective space $\mathbb{R}P^n$ is an $n$-manifold.

We need to show that $\mathbb{R}P^n$ is second countable, locally Euclidean and Hausdorff.
Second countability simply follows from second countability of $\mathbb{R}^{n+1} \setminus \{0\}$.
To prove the locally Euclidean property, I follow a hint and consider the sets $U_i = \{ (x_0,...,x_n) \in \mathbb{R}^{n+1} : x_i \neq 0 \}$. Then we can construct the maps $$\varphi_i : U_i \to \mathbb{R}^n, (x_0,...,x_n) \mapsto \left(\frac{x_0}{x_i},...,\frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i},...,\frac{x_n}{x_i}\right).$$
I don't know how to proceed now. What can we use these maps for? Are they homeomorphisms?
I also haven't been able to prove the Hausdorff property.
Thank you for every hint.

Let $\pi : \mathbb{R}^{n+1} \backslash \{ 0 \} \to \mathbb{R}P^n$ denote the quotient map. In my answer to Real projective space is Hausdorff: is this proof correct? you will find a proof that
a) $\mathbb{R}P^n$ is Hausdorff.
b) The restriction $\hat{\pi} = \pi \mid_{S^n} : S^n \to \mathbb{R}P^n$ is also a quotient map (where $S^n \subset \mathbb{R}^{n+1}$ denotes the standard $n$-sphere).
I guess you know that that $S^n$ is an $n$-manifold. To show that also $\mathbb{R}P^n$ is one, it therefore suffices to show that each sufficiently small $U \subset S^n$ is mapped by $\hat{\pi}$ homeomorphically onto an open $V \subset \mathbb{R}P^n$.
Let $x \in S^n$ and $\varepsilon > 0$. Then $U = \{ y \in S^n \mid \lVert y - x \rVert < \varepsilon \}$ is open in $S^n$ and for sufficiently small $\varepsilon$ we see that $\hat{\pi}$ maps the closure $\overline{U}$ bijectively onto $\hat{\pi}(\overline{U})$. But $\overline{U}$ is compact so that this mapping is a  homeomorphism. Hence also $U$ is mapped homeomorphically onto $\hat{\pi}(U)$. Clearly $\hat{\pi}(U)$ is open
in $\mathbb{R}P^n$ (consider $\hat{\pi}^{-1}(\hat{\pi}(U))$).
