I will use (elementary) results from Lee's Introduction to Smooth Manifolds to streamline the argument.
Let's define the (compact) Mobius band intrinsically (i.e., not via any particular embedding in $\mathbb R^3$) by $M=\mathbb R\times[0,1]/\sim$ where $(x,y)\sim (x+2n,y)$ and $(x,y)\sim(x+2n+1,1-y)$ ($n\in\mathbb N$). Then $M$ has a unique structure which makes the canonical projection map
$$
q:\mathbb R\times [0,1]\to M=\mathbb R^2/\sim,(x,y)\mapsto [x,y],
$$
a local diffeomorphism. Charts in terms of $q$ are then given by $q\vert_U^{-1}$ where $U\subset\mathbb R\times[0,1]$ is an open neighbourhood that doesn't contain distinct points that are identified under the equivalence relation (see also Example 10.3 and Problem 10-1(a) in Lee for the very similar Möbius bundle).
Suppose $M$ is oriented. By Exercise 15.4 in Lee we know that every local frame with connected domain is either positive oriented or negatively oriented. So let's describe certain coordinate frames. To this end, we take specific charts $\phi=q\vert_{(0,1)\times [0,1]}^{-1}$ and $\psi=q\vert_{(1/2,3/2)\times [0,1]}^{-1}$.
Now consider the coordinate frame $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ corresponding to $\phi$. This frame is either positively or negatively oriented (by Exercise 15.4). Likewise, we know that the frame $\frac{\partial}{\partial\tilde x},\frac{\partial}{\partial\tilde y}$ corresponding to $\psi$ is either positively or negatively oriented.
By Proposition 14.6 in Lee, we know that the collection of all oriented smooth charts is a consistently oriented atlas for $M$. It follows that the transition matrix between $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial\tilde x},\frac{\partial}{\partial\tilde y}$ needs to have either positive determinant at each point in $q((0,1/2)\cup (1/2,1)\times [0,1])$ or negative determinant at each point in $q((0,1/2)\cup (1/2,1)\times [0,1])$. Note that the transition map
$$\psi\circ\phi^{-1}:(0,1/2)\cup (1/2,1)\times[0,1]\to (1/2,1)\cup(1,3/2)\times[0,1]
$$
is the identity on $(1/2,1)\times [0,1]$, hence the Jacobian of this map has determinant equal to $1$, which is positive, and therefore it must be positive on $(0,1/2)\cup (1/2,1)\times [0,1]$. However, on $(0,1/2)\times [0,1]$ the transition map $\psi\circ\phi^{-1}$ is given by $(x,y)\mapsto (x+1,1-y)$, whose Jacobian has negative determinant. We've arrived at a contradiction, and therefore $M$ is not orientable.