Let's say that $\mathcal{M}$ is a smooth submanifold of dimension $n$, of $\mathbb{R}^{n}$.
Using my definition, this means: for every point $p \in \mathcal{M}$ there exists a coordinate patch $\alpha: U \to V\cap\mathcal{M}$, where $U,V\subseteq\mathbb{R}^{n}$ are both open, and $p \in V$ such that $\alpha$ is a smooth homeomorphism, and where $\mathcal{J}_{\alpha}$ (Jacobian matrix) has rank $n$ on all of $U$.
My definition of a natural orientation: a collection of coordinate charts $\{\alpha_{j} : U_{j} \to V_{j} \}_{j \in J}$ such that $\det\left( \mathcal{D}_{\alpha_{j}}(x) \right)>0$ for all $x \in U_{j}$, for all $j \in J$.
I'm trying to show that $\mathcal{M}$ always has a natural orientation.
$\ $
My attempt: I'm thinking, I should take a smooth atlas $\{\alpha_{j} : U_{j} \to V_{j} \}_{j \in J}$ of $\mathcal{M}$. If I am able to show that $\det\left( \mathcal{D}_{\alpha_{j}}(x)\right) \neq 0$ on all of $U_{j}$, then by the intermediate value theorem this means that either $\det\left( \mathcal{D}_{\alpha_{j}}(x)\right) > 0$ or $\det\left( \mathcal{D}_{\alpha_{j}}(x)\right) < 0$ on all of $U_{j}$.
In the case that its positive, I keep $\alpha_{j}$ in the atlas. In the case that it's negative, then I replace $\alpha_{j}$ with $\beta_{j}$ in the atlas, where $\beta_{j} = \alpha_{j} \circ r$ such that $r$ is a homeomorphism with negative Jacobian determinant (like $r(x_{1}, \ldots, x_{n})=(-x_{1}, \ldots x_{n})$ for example). The new atlas is now a natural orientation.
So my proof hinges on the fact that the determinant of the Jacobian of a coordinate patch on $\mathcal{M}$ is never zero. Is this true?