I would like to prove that two immersions $f,g: S^1 \to \mathbb{R}^2$ where $f$ is a figure eight and $g$ is a circle are not regularly homotopic.
I'm pretty sure this can be done by noting that a figure eight has an odd number of crossings, whereas the circle has an even number of crossings. There is no sequence of Reidemeister moves II and III transforming the knots to the other one. Thus, the two immersions are nor homotopic.
But is there a way to do this without referencing results from knot theory?
For any immersion $f:S^1 \to \Bbb R^2$, there is a canonical map $\tilde f:S^1\to S^1$ with $\theta \mapsto df(\theta)/||df(\theta)||$. If $f$ and $g$ are regularly homotopic, it is easy to see that $\tilde f$ is homotopic to $\tilde g$. Define $\gamma(f)$ (the rotation number) to be the degree of $\tilde f$ (picking the usual counter-clockwise orientation).
For $f$ the figure eight and $g$ a round circle, the regular value definition of degree implies that $\gamma(f)=0$ and $\gamma (g) =\pm 1$ (depending on how you orient your circle).
As it turns out, $\gamma$ is actually a complete invariant. The proof of this is rather simple and exposited well in Hassler Whitney's original article On Regular Closed Curves in the Plane. One beautiful book is Arnol'd's Topological Invariants of Plane Curves and Caustics.