When is an open subset of an orientable manifold an orientable submanifold?

I am trying to complete problem $ 4$ from the introductory chapter of Riemannian Geometry by do Carmo. The question asks us to show that the projective plane $ P^2(\mathbb{R})$ is non-orientable, first by proving any open subset of an orientable manifold is an orientable submanifold.

do Carmo’s definition for a manifold $ M$ to be orientable is if it admits a differentiable structure $ \left\{(U_\alpha,\mathbf{x_\alpha}) \right\}_{\alpha \in \Gamma}$ such that for every pair $ \alpha,\beta$ with $ \mathbf{x_\alpha}(U_\alpha)\cap\mathbf{x_\beta}(U_\beta)$ nonempty the differential of the change of coordinates, $ d(\mathbf{x_\beta}^{-1} \circ \mathbf{x_\alpha})$ , has positive determinant.

My approach to this was to say let open $ V \subset M$ be given. Then $ \left\{(U_\alpha\cap \mathbf{x_\alpha}^{-1}(V),\mathbf{x_\alpha}) \right\}_{\alpha \in \Gamma}$ is a differentiable structure on $ V$ that satisfies the definition.

My confusion stems from the fact that an open Möbius band can embedded in $ \mathbb{R}^3$ , for example by the embedding $ f:U\to \mathbb{R}^3$

$ $ f(x,y) = \left(\left(1+\frac{y}{2}\cos\frac{x}{2}\right)\cos{x},\;\left(1+\frac{y}{2}\cos\frac{x}{2}\right)\sin{x},\;\frac{y}{2}\sin\frac{x}{2}\right)$ $

Where $ U = \left\{(x,y)\in \mathbb{R}^2 : 0 \leq x < 2\pi,\;-1 < y < 1\right\}$ (taken from Wikipedia).

So I guess my confusion comes down to “why isn’t the open Möbius band a submanifold of $ \mathbb{R}^3$ ?”