let $f:(0,1] \rightarrow \mathbb{R^2} , f(t)=\left(t \cos \left( \frac{1}{t}\right), t \sin \left(\frac{1}{t}\right)\right )$
is $f$ an immersion, embedding or proper embedding?
$f$ is an immersion because $f^\prime(t)\neq 0$ for all $t \in (0,1]$ and $f$ is one to one.for embedding we must show $f:(0,1] \rightarrow im(f)$ is open map .
Let $D^2 = \{(x,y) \in \mathbb R^2 \mid \lVert (x,y) \Vert \le 1\}$ denote the closed unit disk and define $n : D^2 \setminus \{0\} \to (0,1], n(x,y) = \lVert (x,y) \rVert$.
We have $\lVert f(t) \rVert =\sqrt{(t \cos( \frac{1}{t}))^2, (t \sin (\frac{1}{t})^2} = t \in (0,1]$, thus $f((0,1]) \subset D^2 \setminus \{0\}$. Then $n(f(t)) = t$ which shows that $f$ is injective (if $f(t) = f(t')$, then $t = n(f(t)) = n(f(t')) = t'$).
Thus $\bar f : (0,1] \stackrel{f}{\to} f((0,1])$ is a continuous bijection; we want to show that it is a homeomorphism (which is equivalent to $\bar f$ being an open map). Let $g = n \mid_{f((0,1])} : f((0,1]) \to (0,1]$. This is a continuous map such that $g \circ \bar f = id_{(0,1]}$ Let $\bar f^{-1} : f((0,1]) \to (0,1]$ denote the inverse of $\bar f$ (at the moment we only know that is a function between sets). Then $\bar f^{-1} = id_{(0,1]} \circ \bar f^{-1} = (g \circ \bar f) \circ f^{-1} = g \circ (\bar f \circ f^{-1}) = g \circ id_{f((0,1])} = g$ which means that $\bar f^{-1}$ is continuous.