An injective immersion that is not a topological embedding On page 86 of John Lee's Introduction to smooth manifolds there is an example of an injective immersion that is not a topological embedding:

$\beta : (-\pi, \pi) \to \mathbb{R}^2$, defined by $\beta(t) = (\sin{2t}, \sin{t})$, or pictorially:



It is explained that, although $\beta$ is an injective immersion, it is not a smooth embedding since the image is compact while the domain is not.
My understanding is that the image, while bounded in $\mathbb{R}^2$, is an open subset of the plane, whereas the statement claims that it is not.

Would anyone please explain why the image is compact? Thank you.
First proof: If $\beta(t_n)$ is a sequence of points in the image, the sequence $t_n$ is bounded in $\mathbb{R}$, hence there is a subsequence $t_{n_p}$ that converges to a $t \in [-\pi, +\pi]$. By continuity of the sine, $\beta(t_{n_p})$ converges to $(\sin(2t), \sin(t))$, which is equal to $\beta(t)$ if $t\in (-\pi, \pi)$ and to $\beta(0)$ otherwise. Thus, every sequence in the image has a subsequence that converges in the image, which is the definition of compactness.

Second proof: Let $\gamma$ be the map $t \mapsto (\sin(2t), \sin(t))$ from $[-\pi, \pi]$ into $\mathbb{R}^2$ The image of $\gamma$ is the same as $\beta$, hence it is the image of a compact set by a continuous map.
I would like to add a simple explanation. We can just use the fact that $\beta \left ( -\pi , \pi \right )$ is bounded (and closed because it has all its limit points) in $\mathbb{R}^{2}$ therefore it's compact in the topology of $\mathbb{R}^{2}$. It is a simple exercise to check that compactness in the bigger space ($\mathbb{R}^{2}$) implies compactness in the subspace topology of $\beta \left ( -\pi , \pi \right )$. Thus $\beta \left ( -\pi , \pi \right )$ is a compact space and also an image of a non-compact set. This shows that the map can not be a homeomorphism.