The Real Projective Plane is a 2-dimensional manifold To demonstrate this do I just have to show that $P^2$ is Hausdorff and locally Euclidean? I can show that the space is Hausdorff but I'm having a little trouble demonstrating that it is locally Euclidean. 
 A: The real projective plane, $\mathbb{R}P^2$, is the space of lines through the origin in $\mathbb{R}^3$, so it's natural to use homogeneous coordinates, where $(x, y, z) \sim (tx, ty, tz)$ for any $t \in \mathbb{R} \setminus \{0\}$.
We can manage to cover the projective plane with three charts, one for each coordinate in $\mathbb{R}^3$.  Here's one of them.  Let $U_x \subset \mathbb{R}P^2$ denote the subset of points in $\mathbb{R}^3$ with nonzero $x$-coordinate.
$$U_x = \{ (x, y, z) \in \mathbb{R}^3 \; | \; x \ne 0\}$$
Considering these as homogeneous coordinates, though,
$$(x, y, z) \sim \left(1, \frac{y}{x}, \frac{z}{x}\right).$$
Geometrically, this projects any line that pierces the plane $x = 1$ onto its intersection point with that plane.  Voilà, this is our first chart.
$$\begin{align} \psi_x: U_x &\to \mathbb{R}^2 \\ (x,y,z) &\mapsto \left(\frac{y}{x}, \frac{z}{x}\right) \end{align}$$
Notice that the inverse, $\psi_x^{-1}: \mathbb{R}^2 \to U_x$ is given by $(u, v) \mapsto (1, u, v)$.
In a strictly analogous fashion, we can construct charts for open sets $U_y$ and $U_z$.  To stitch these charts together into an atlas, we have to look at the various compositions that arise where the charts overlap, mapping open subsets of Euclidean space into itself.
$$\begin{array}{*{5}{c}}
\psi_x(U_x \cap U_y) &\to& U_x \cap U_y &\to& \psi_y(U_x \cap U_y) \\
(u, v) &\mapsto& (1, u, v) &\mapsto& \left( \frac{1}{u}, \frac{v}{u}\right)
\end{array}$$
This composition is a diffeomorphism, which is to say that its smooth and so is its inverse.  This is what makes $\mathbb{R}P^2$ into a smooth manifold.  (It may be that you're only interested in the projective plane as a topological manifold, in which case you are only interested in these maps being homeomorphisms.)
