Old Standby 2: $\mathbb{R}P^n$ is orientable iff $n$ is odd.

First note that the antipodal map $a:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ by $x\mapsto -x$ is orientation preserving if n is odd and orientation reversing if n is even just because in coordinates it is the matrix with $-1$‘s on the diagonal.

Now if we make our natural identifications between $\mathbb{R}^{n+1}$ as a manifold and as the vector space that is tangent space at a given point, then we see that if we restrict $a$ to $S^n$ embedded in $\mathbb{R}^{n+1}$ the orientation preserving/reversing still holds. This is just because if $(v_1, \ldots, v_n)$ is an oriented basis at $p\in S^n$, then $(p, v_1, \ldots, v_n)$ is an oriented basis at that same point in $\mathbb{R}^{n+1}$. Thus the orientation at $-p$ of $a(p, v_1, \ldots , v_n)=(-p, -v_1, \ldots, -v_n)$ is $(-v_1, \ldots, -v_n)$.

Now suppose that n is even and that $\mathbb{R}P^n$ has an orientation. Let $\pi: S^n\to\mathbb{R}P^n$ be the standard quotient map. But now the orientation of $\mathbb{R}P^n$ induces an orientation on $S^n$ by letting an ordered basis $(v_1, \ldots , v_n)\subset T_pS^n$ be positively oriented if $(d\pi_p(v_1), \ldots, d\pi_p(v_n))$ is positively oriented.

But the induced map of $a$ on $\mathbb{R}P^n$ is just the identity. Thus $a$ is orientation preserving on $S^n$, a contradiction.

The other direction we need to put an orientation on $\mathbb{R}P^n$ by $S^n$. Suppose n is odd now. Define a basis $(w_1, \ldots , w_n)\subset T_{\pi(p)}\mathbb{R}P^n$ to be positively if there exists a positively oriented basis $(v_1, \ldots , v_n)\subset T_pS^n$ such that $(d\pi_p(v_1), \ldots , d\pi_p(v_n))=(w_1, \ldots , w_n)$. We just need to make sure this is a well-defined choice. But it is since the fibers of $\pi(p)$ are $p$ and $-p$, and we get from one to the other through the antipodal map which is orientation preserving. So we’re done.

Let’s do some analysis of this. First off, there was nothing special about this particular $\pi$. So we actually proved that if $\pi: N\to M$ is a smooth covering and $M$ is orientable, then $N$ is also orientable.

I know of two other ways to prove this. Both require that antipodal map observation first. One way is to prove the more general fact that if $M$ is a connected, oriented, smooth manifold and $G$ is a discrete group acting smoothly, freely, and properly on M, then $M/G$ is orientable iff $x\mapsto g\cdot x$ is an orientation preserving diffeo for all $g\in G$. In this case, $G\{\pm 1\}$ and the 1 action is the identity and the $-1$ action is the antipodal map.

The other way is far more elegant. It is the algebraic topology method (actually I believe you can prove the stronger statement of the second method using this way). I haven’t quite reworked this way out, yet, but it just involves pulling back nowhere vanishing n-forms or something.