Embedded submanifolds: the circle into $\mathbb R^2$. I struggle to wrap my head around homeomorphisms and embeddings.
Especially I considered the circle and how to embed it into $\mathbb{R}^2$.
Consider the function $f: [0,2\pi) \rightarrow \mathbb{R}^2; x \mapsto (\cos x,\sin x)$. This is no homeomorphism since $f^{-1}$ is not continuous. Another way of seeing that no homeomorphism exists can be seen by observing $[0,2\pi)$ is open but $f([0,2\pi))$ is nether open nor closed. [Wikipedia/Homeomorphism][1] and [Stackoverflow][2]
Is this deduction correct?
The real confusion begins now.
Because $f$ is not a homeomorphism, it cannot be an embedding, which are homeomorphisms onto their images.([Wikipedia/Homeomorphism][3])
However, some authors argue the opposite; see, for example, [Link, Example 3.2 end] [4].
Thus, I suppose I misunderstood something.
If I'm correct that $f$ is no embedding of the unit circle onto $\mathbb{R}^2$ how can it be shown that the circle can be embedded into $\mathbb{R}^2$? Or more general how can one show that $S^{n-1}$ (the (n-1)-dimensional unit sphere) can be embedded into $\mathbb{R}^n$?
  [1]: https://en.wikipedia.org/wiki/Homeomorphism#Notes
[2]: Wrongly showing that an interval has a homeomorphism to $S^1$
  [3]: https://en.wikipedia.org/wiki/Embedding#General_topology
  [4]: https://math.uchicago.edu/~may/REU2019/REUPapers/Smith,Zoe.pdf
 A: Perhaps you should start with reading precisely what a homeomorphism and embedding are:

*

*A function $f:X\to Y$ is a homeomorphism if it is invertible, continuous and $f^{-1}$ is continuous.

*Typically, a function $f:X\to Y$ is called an embedding if $f$, as a function $X\to f(X)$, is a homeomorphism.

Note that evey homeomorphism is an embedding, but not vice versa.
It is important for you to understand the role of $Y$ here. If $f$ is not "onto", then it is not invertible, and thus not a homeomorphism. It still can be an embedding though.

Consider the function $f: [0,2\pi) \rightarrow \mathbb{R}^2; x \mapsto (\cos x,\sin x)$. This is no homeomorphism since $f^{-1}$ is not continuous.

Incorrect. This is not a homeomorphism because it is not invertible: forget about $f^{-1}$ being continuous, it doesn't exist! Because the image is not whole $\mathbb{R}^2$, $f$ is not "onto".
And it is not an embedding, because its image is $S^1$, and $f$ (as a function $[0,2\pi)\to S^1$) indeed does not have a continuous inverse.
And in fact if $f:[0,2\pi)\to \mathbb{R}^2$ with $S^1$ as its image, then $f$ is not an embedding, regardless of how $f$ is defined. Because then it would induce a homeomorphism $[0,2\pi)\to S^1$, but $S^1$ is compact while $[0,2\pi)$ is not.

The real confusion begins now. Because $f$ is not a homeomorphism, it cannot be an embedding, which are homeomorphisms onto their images.

Homeomorphism onto image is not a homeomorphism. Embedding is not a homeomorphism in general. I mean, read again your claim carefuly. You say "not homeomorphism $\Rightarrow$ not embedding", or equivalently "embedding $\Rightarrow$ homeomorphism". And so is the term "embedding" and "homeomorphism" the same? Of course not. But yeah, I understand how this naming convention can be confusing.

how can it be shown that the circle can be embedded into $\mathbb{R}^2$?

Well, the circle, or any sphere is defined as
$$S^n=\big\{v\in\mathbb{R}^{n+1}\ \big|\ \lVert v\rVert =1\big\}$$
and so there is one obvious candidate: the inclusion
$$i:S^n\to\mathbb{R}^{n+1}$$
$$i(x)=x$$
It is an embedding, because the image is $S^1$ and $i^{-1}=i$ (on image).
It is not a homeomorphism, because $i$, as a function $S^n\to\mathbb{R}^{n+1}$ is not invertible.
