Why isn't $\mathbb{RP}^2$ orientable?

How to prove that $\mathbb{RP}^2$ isn't orientable? My book (do Carmo "Riemannian Geometry") gives a hint: "Show that it has a open subset diffeomorphic to the Möbius band", but I don't know even who is the "open subset".

For geometric understanding of the real projective plane, I prefer to think of it as gotten from a closed circular disk by identifying opposite points on the boundary. If you accept this, then a Möbius band subset is easily found: take any diameter of your original circle, and widen it to a strip.

$\mathbb{RP}^2$ is the 2-sphere, modulo the relation identifying opposite points. The open subset referred to is a neighborhood of the equator on the sphere. Consider the tangent vector at the equator pointing upwards, and rotate the sphere horizontally by $\pi$. Using the quotient map from the sphere to $\mathbb{RP}^2$, you will see that your tangent vector is in the same fiber as your original tangent vector, but which way will the tangent vector point now?

Here are some general facts which will help us to solve the problem in more general level. 1. The Antipodal map $a:S^n\longrightarrow S^n$ on the n-sphere is defined by $$a(x^1,\dots,x^{n+1})=(-x^1,\dots,-x^{n+1})$$ is orientation preserving map iff n is odd. 2.Suppose $f(x^1,\dots,x^{n+1})$ is a $\mathbb{C}^{\infty}$ function on $\mathbb{R}^{n+1}$ with 0 as a regular value. Then the zero set of $f$ is an orientable submanifold of $\mathbb{R}^{n+1}$, In particular $S^n$ is orientable in $\mathbb{R}^{n+1}$ So use these facts and you will get any odd dimensional projective space is orientable.