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.
 A: The image is literally a leminscate in $\Bbb R^2$. 
It's clearly not open as if you take a point on the leminscate, any small neighborhood of it in $\Bbb R^2$ gets outside the curve (i.e. hits the complement). It's in fact closed, because a leminscate is a level curve which are closed because they are preimage of $0 \in \Bbb R$ by a continuous function.
As you noted, it's bounded, so that guarantees compactness.
A: 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.
A: It contains all its limit points, so it is a closed subset of $\mathbb{R}^2$. Since it is bounded as well, by the Heine-Borel theorem, it is a compact subset of the plane $\mathbb{R}^2$.
A: The image is a solid figure eight: the only point of contention is the origin, but the "hole" between the abutting open ends is "plugged" by the middle of the curve.
A: Note that interior of $\beta ((-\pi,\pi)) \subset \mathbb{R}^2$ is empty. So it cannot be an open subset of $\mathbb{R}^2$. Now take any open cover $\{U_i\}_{i \in I}$ of the image. For some $j \in I$, we must have  $0 \in U_j$. Clearly, the rest of the image can be covered by finitely many $U_k$'s where $k \in I$. So we started with any open cover and find a finite subcover. That is, image is compact subset of $\mathbb{R}^2$.
