Embedding $ \mathbb{R} \to \mathbb{R}^2$ The question I'm looking at is as follows:

Prove that there is an embedding of the line as a closed subset of the plane, and there is an embedding of the line as a bounded subset of the plane, but there is no embedding of the line as a closed and bounded subset of the plane.

My understanding of embedding is that it needs to be a homomorphism from $ \mathbb{R} \to f(x) \in \mathbb{R}^2$. I.e. the entire number line needs to be in $ \mathbb{R}^2$ in some shape or form after the transformation.
My thoughts for the closed subset are simply $ f(x) : x \to (1,x) $ as this is is effectively the identity function plus one dimension. It is closed as all 0 limits are contained, and unbounded as the Cauchy sequences do not converge as x approaches $ - \infty $ and $ \infty $.
For bounded $ f(x) : x \to (arctanh(x),x) $ on $ (-1,1) $ which encodes the entire number line, has Cauchy sequences converging at limits, but does not contain $ x = -1 $ or $ x = 1 $.
Are these intuitions correct for these parts of the question, or am I misinterpreting embedding as a concept? Are there significantly simpler answers? I feel like I'm missing something.
 A: The first example is fine (essentially embed the reals as a straight line in the plane), which gives us an embedding $e$ of $\mathbb{R}$ into $\mathbb{R}^2$ where $e[\mathbb{R}]$ is closed.
To make it bounded, use the essentially same idea, using that $f(x) = \arctan(x)$ is a homeomorphism between $\mathbb{R}$ and $(-\frac{\pi}{2},\frac{\pi}{2})$ which is a bounded open interval. To get it into the plane, add a fixed coordinate: $e(x) = (\arctan(x),0)$, which has the bounded image $(-\frac{\pi}{2},\frac{\pi}{2}) \times \{0\}$ and is still a homeomorphism between $\mathbb{R}$ and its image. Adding $x$ as a second coordinate makes it unbounded again (it is a homeomorphism still).
If there were some embedding $e: \mathbb{R} \rightarrow \mathbb{R}^2$ where $e[\mathbb{R}]$ (which is homeomorphic to $\mathbb{R}$ by definition of an embedding!) would be closed and bounded, and $e[\mathbb{R}]$ would be compact by the Heine-Borel theorem, which $\mathbb{R}$ is not (and compactness is preserved by homeomorphism...).
