I'm currently stuck trying to demonstrate that this is both second-countable and Hausdorff. First, my book defines $\mathbb R \mathbb P^n$ as the set of 1-dimensional linear subspaces of $R^{n+1}$ wit the quotient topology determined by the natural map $\pi:\, \mathbb{R}^{n+1}\setminus \{0\} \to \mathbb R \mathbb P^n$ sending each point $x \in \mathbb{R}^{n+1}$ to the subspace spanned by $x$.
The definition leads me to believe that $\pi(x)=kx$ for $x\in\mathbb{R}.$ Furthermore, $\pi$ is a homeomorphism.
The preceding example makes a mention that $\mathbb R^{n+1}$ is second-countable, and a subspace of a second-countable topological space is second-countable.
However, I'm not sure how to argue second-countability without the heavy machinery of the second countability theorem (which I have been reassured is not necessary for this problem).
Edit: Let $U \subseteq R^{n+1}\setminus \{0\}$ be an open set. Then $\lambda U$ is also an open set in $R^{n+1}\setminus\{0\}.$ I'm not sure how the following works: $\pi^{-1}(\pi(\cup_{\lambda}\lambda U))$ is an open set, so $\pi(U)$ is an open set in $\mathbb R \mathbb P^n$.
Maybe the best way to do this is to break your quotient map into two steps. First quotient out by the relation $x \sim rx$ for $r \in \mathbb{R}^+$, to get the sphere $S^n$, which is a manifold (you can either explicitly make finitely many charts, or use the implicit function theorem on $x_1^2+x_2^2+\dots+x_n^2-1$). The Hausdorff and second countability conditions follow from the fact that it is topologically embedded in $\mathbb{R}^{n+1}$, which is both Hausdorff and second countable.
Now what remains to be identified are the antipodal points. This quotient map $\pi:S^n \to \mathbb{RP}^n$ is very nice (in fact it is a covering map). In particular, the map is a surjective local homeomorphism, so $\mathbb{RP}^n$ is a manifold by pushing forward local charts for $S^n$. Since the map is open, $\mathbb{RP}^n$ is second countable, and is also Hausdorff since to separate points on $\mathbb{RP}^n$, one must separate two distinct pairs of antipodal points on $S^n$, but $S^n$ is normal.