Define the figure-8 curve $S$ to be the image of the smooth map $\beta : (-\pi,\pi) → \mathbb{R}^2$ given by $\beta(t)=(\sin 2t, \sin t)$.
I am trying to prove that $S$ is not an embedded (or regular) submanifold of $\mathbb{R}^2$.
The thing is, I am teaching myself differential geometry, and the chapter on submanifolds has been very difficult, so I would appreciate it if someone verified that I was thinking correctly about this stuff.
Here are my proofs. I believe the 2nd proof is the standard approach.
Proof 1: First of all, note that $S$ is an immersed submanifold, with the submanifold topology inherited from the immersion $\beta$. If $S$ were also an embedded submanifold of $\mathbb{R}^2$, then because $\beta$ is smooth, we may restrict the codomain to $S$ and conclude that $\beta$ is also smooth as a map onto its image $S$, but $S$ is not even continuous with its submanifold topolgy, let alone smooth. This is a contradiction, so $S$ cannot be an embedded submanifold of $\mathbb{R}^2$.
Proof 2 (sketch): By assuming $S$ is an embedded submanifold, we may use the local slice condition to arrive at a contradiction by considering the number of components, which ought to be preserved by homeomorphisms but isn't here.
Edit: So it turns out that Proof 1 does not work (see comments). I might try reworking the argument and editing the question with it, but ultimately, it seems only Proof 2 is correct.