Lemniscate is homeomorphic to the circle? Do you agree that a lemniscate (i.e. $\infty$) is homeomorphic to the circle if theses two curves are embedding in $\mathbb R^3$, but that this result is not true in $\mathbb R^2$ ?
I read somewhere that the lemniscate is never not homeomorphic to the circle, but I have doubt about this information (without context). So is my example correct ? 
 A: They are of course not homeomorphic, but in $\mathbb R^3$, you have that $$\mathbb S^1\cong L/_{\sim}$$
where $L$ is the Lemniscate, and $\sim$ a relation of equivalence (the one that "glue the two circles of the lemniscate"... strange expression, but it's in fact very natural. Tell me if you want a draw). Now in $\mathbb R^2$ you can get the same bijection (between $L/_{\sim}$ and $\mathbb S^1$), but it will not be continuous. Therefore, it's not an homeomorphism (and there is no way to "glue continuously the two circle of the lemniscate to get a circle"). Therefore, they will not be homeomorphic.
I'm trying to be simple in my words, I hope it will help. Tell me if it doesn't.
A: What you read is correct. The lemniscate and and the circle are intrinsically non-homeomorphic, regardless in what space they are embedded. 
The lemniscate has a cut-point (if we remove it, the remainder (two open intervals, essentially, is not connected) and the circle has not, and this will always result in the spaces not being homeomorphic. 
