I can't furnish a formal proof, but perhaps this animation can help you "see" why the two bands must be interlinked. Notice how the "marginal" purple band essentially goes through the loop created by the central band.
I'm not a topologist but upon further reflection, can offer the following idea. In the parametrization $$\vec m (u,v) = \begin{bmatrix} (1 + u \cos \tfrac{v}{2}) \cos v \\ (1 + u \cos \tfrac{v}{2}) \sin v \\ u \sin \tfrac{v}{2} \end{bmatrix}$$ for $(u,v) \in [-1,1] \times (-\pi, \pi]$, consider the curves $$\gamma_1 = \vec m(0,v), \quad v \in (-\pi, \pi]$$ which is the unit circle in the $xy$-plane, and $$\gamma_2 = \vec m(1,v), \quad v \in (-2\pi, 2\pi]$$ which is the boundary of the Mobius band. Since the central portion of the band contains $\gamma_1$, and the purple "marginal" band contains $\gamma_2$, then the latter must be interlinked with the former if $\gamma_2$ intersects the $xy$-plane in exactly two points, once inside the unit disk, and once outside. To this end, we solve for the cases where $z = 0$ for $\gamma_2$: i.e., $\sin \frac{v}{2} = 0$, or $v = 2\pi k$ for some integer $k$. Thus the intersection points are $(0,0,0)$ for $k$ odd, and $(2,0,0)$ for $k$ even, and since these are unique, it follows that the purple band is interlinked with the blue.