William Boothby, An Introduction to Differentiable Manifolds and Riemannian Geometry Rev 2nd ed. At the beginning of ch3. You can find a detailed proof that $\mathbb RP^n$ admits a differentiable manifold structure over the quotient topology induced by the natural projection $\pi:\mathbb R^{n+1}\to\mathbb RP^n$.
To see it is compact and connected, this is really a topological matter. Restrict the domain of $\pi$ to $S^n$. According to the definition of quotient topology, $\pi:\mathbb R^{n+1}\to\mathbb RP^n$ is a surjective continuous map. So $\pi|_{S^n}:S^n\to\mathbb RP^n$ with $S^n$ equipped with subspace topology is also a surjective continuous map. Surjectivity of $\pi|_{S^n}$ comes from the fact that $\forall [x_0,x_1,\dots,x_n]\in\mathbb RP^n$ has in its preimage a point $(x_0,\dots,x_n)/(\sum_{i=0}^nx_i^2)^{1/2}\in S^n$. Therefore. it preserves compactness and connectedness. So $\mathbb RP^n$ is compact and connected since $S^n$ is.
Appendix:
- $S^n$ is compact, since it is closed and bounded in $\mathbb R^{n+1}$ by Heine-Borel theorem.
- $S^n$ is path connected: any two points can be connected by an arc on a great circle. So it is also connected. (path connectedness implies connectedness).
- All the topological facts can be picked up in James Munkres's Topology 2nd ed. Or you can find a quick review in Abraham, Marsden, Ratiu, Manifolds, Tensor Analysis and Applications (2007 draft recommended).