No solenoid can be embedded into the Mobius strip. To derive a contradiction, assume that some solenoid $S$ embeds into the Mobius strip $M$. Let $\pi:C\to M$ be a 2-fold covering map of the cylinder $C$ onto the Mobius strip. It is well-known that the solenoid $S$ contains a dense subset $D$ which is the image of the real line under a continuous map $\phi:\mathbb R\to D$ (and this image of $\mathbb R$ is called a *composant* of the solenoid). By the lifting property of the covering map $\pi$, there exists a continuous map $\varphi:\mathbb R\to C$ such that $\pi\circ\varphi=\phi$. Then the closure $K$ of the connected set $\phi(\mathbb R)$ in $C$ is a continuum such that $\pi(K)=\bar D=S$. Taking into account that the cylinder $C$ embeds into the plane, we conclude that $K$ is a planar continuum and hence the solenoid $S$ is a continuous image of a planar continuum.

On the other hand, by a result of Krasinkiewicz in his paper Mappings onto circle-like continua, no solenoid is a continuous image of a planar continuum (as solenoids have infinitely divisible first cohomology group whereas the first cohomology group of any planar continuum is finitely divisible). This is a desired contradiction.