My ultimate goal is to show that the real projective space $\mathbb{P}^n_{\mathbb{R}}$ is an $n$-manifold. But first I'd like to understand the topological structure of $\mathbb{P}^n_{\mathbb{R}}$.
The quotient topology on $\mathbb{P}^n_{\mathbb{R}}$ is defined via the quotient map $q \colon \mathbb{R}^{n+1} \backslash \{0\} \to \mathbb{P}^n_{\mathbb{R}}$ sending $x \mapsto [x]$, its linear span. Thus open subsets of $\mathbb{P}^n_{\mathbb{R}}$ are collections of lines through the origin in $\mathbb{R}^{n+1}$ (correct)? Then $\beta := \{U_i\}_{0 \le i \le n}$ is a basis for the topology on $\mathbb{P}^n_{\mathbb{R}}$ (correct?), where $U_i = \{[(x_0, \ldots, x_n)]: x_i = 1 \}$ since each $U_i$ contains every line through the origin for a fixed $i$; so $\mathbb{P}^n_{\mathbb{R}}$ is second countable.
Proving $\mathbb{P}^n_{\mathbb{R}}$ is Hausdorff is where I am having a bit of trouble. I understand that this should be true since distinct points in $\mathbb{P}^n_{\mathbb{R}}$ are just distinct lines in $\mathbb{R}^{n+1}$, but I cannot think of a way to say this precisely. What is an open neighborhood of a point in $\mathbb{P}^n_{\mathbb{R}}$? I think it is the line itself with a "bundle" of the lines contained in a circle around it. But again, I am unsure of how to make this precise. This is my attempt: let $[v]$ and $[w]$ be distinct points in $\mathbb{P}^n_{\mathbb{R}}$. Let $d(v,w) = \delta$ (the Euclidean metric on $\mathbb{R}^{n+1}$), then $U = \{[u]: d(u,v) < \frac{\delta}{2} \}$ and $Z = \{[z] : d(z,w) < \frac{\delta}{2}\}$ are disjoint open neighborhoods of $[v]$ and $[w]$, respectively.
I think I am okay with the locally Euclidean property.
Suggestions/corrections would be appreciated.
As a lemma, we show that $q$ is an open quotient. If suffices to show that if $B\subset\mathbb R^{n+1}\setminus\{0\}$ is an open ball, then $q(B)$ is open. We know that $q^{-1}\circ q(B)=C$ is the set of all origin-less linear subspaces intersecting nontrivially with $B$. Pick a point $x\in C$. Then $x=\lambda y$ for some $x\in B$. Since $B$ is open, so is $\lambda B\subset C$, and $y\in \lambda B$, so $C$ is open. Thus, $q(B)$ is also open, and $q$ is an open map.
Now we show that $\mathbb{RP}^n$ is Hausdorff. Let $x,y\in\mathbb{RP}^n$ be distinct. Choose $x'$ and $y'$ as points in $q^{-1}(\{x\})$ and $q^{-1}(\{y\})$ respectively. Let $L_1$ and $L_2$ be the lines through the origin containing $x'$ and $y'$ respectively. We know that neither $x'$ nor $y'$ are the origin, and that $L_1$ and $L_2$ are distinct, intersecting only at the origin. Let $$C_1=\{z\in L_1:|z|\geq |x'|\}$$ and $$C_2=\{z\in L_2:|z|\geq |y'|\}.$$ Then $C_1$ and $L_2$ are disjoint closed subspaces of $\mathbb R^{n+1}$. Since $\mathbb R^{n+1}$ is normal, there exist disjoint open sets $U_1$ and $V_2$ containing $C_1$ and $L_2$ respectively. Likewise there exist open sets $V_1$ and $U_2$ containing $L_1$ and $C_2$ respectively. Thus, $U=U_1\cap U_2$ and $V=V_1\cap V_2$ are disjoint open sets contain $C_1$ and $C_2$, with $U$ disjoint from $L_2$ and $V$ disjoint from $L_1$. Also note that neither $U$ nor $V$ contains teh origin, so we may think of these sets as being open sets in $\mathbb R^{n+1}\setminus \{0\}$. Therefore, $q(U)$ and $q(V)$ are disjoint open sets containing $x$ and $y$.