EDIT: Due to some confusion in the comments as to what my question is, note the following is an almost (modified to dimension 1) direct quote from Leon Simon's book on Geometric Measure Theory:
Let $M$ be a 1 dimensional $C^1$ submanifold of $\mathbb{R}^2$. In other words, $M \subset \mathbb{R}^2$, and for each $(x,y) \in M$, there are open sets $V \subset \mathbb{R}$, $W \subset \mathbb{R}^2$, with $(x,y) \in W$ and a $1-1$, $C^1$ map $\psi: V \to W$ with
$$ \psi(V) = W \cap M$$
where $\psi $ is proper (if $K\subset W$ compact, then $\psi^{-1} K$ is compact in $V$.), and $D \psi(x,y) \neq 0$ for each $(x,y) \in V$.
The condition that $\psi$ is proper includes examples like
$$M = \{ (x, \sin 1/x): x >0 \}$$
but eliminates examples such as
$$M = \{ 0 \} \times (-1,1) \cup \{ (x, \sin 1/x): x >0 \}.$$
END QUOTE
Note no $\psi$ is specified.
Question: Why is there a proper $\psi$ for $$M = \{ (x, \sin 1/x): x >0 \}$$ but there is no proper $\psi$ for
$$M = \{ 0 \} \times (-1,1) \cup \{ (x, \sin 1/x): x >0 \}?$$
I understand that the second example is not a submanifold because its not pathconnected, and homeomorphisms preserve path-connectedness. What does $\psi$ being proper have to do with it?
For $M = \{ (x, \sin 1/x): x >0 \}$ and all points of $M$ we can take the proper map $$\psi : (0,\infty) \to (0,\infty) \times \mathbb R, \psi(x) = (x, \sin 1/x) .$$
For $M' = \{ 0 \} \times (-1,1) \cup M$ consider any point $p \in M'$. We want to show that there are open sets $V \subset \mathbb{R}$, $W \subset \mathbb{R}^2$ with $p \in W$ and an injective $C^1$ map $\psi: V \to W$ such that $\psi(V) = W \cap M'$ (we do not require $\psi$ to be proper).
Define $W = \mathbb R^2$ (which is an open neighborhood of $p$) and $V = (-2,0) \cup (0,\infty)$ (which is an open subset of $\mathbb R$). The map $$\psi : V \to W, \psi(x) = \begin{cases} (0,x+1) & x \in (-2,0) \\ (x, \sin 1/x) & x \in (0,\infty) \end{cases}$$ is an injective $C^1$ map such that $\psi(V) = M' = W \cap M'$.
It is not proper.
Consider the compact set $K = \{(0,0)\} \cup \{(1/ n\pi,0) : n \in \mathbb N \}$. Then $\psi^{-1}(K) = \{-1\} \cup \{1/ n\pi : n \in \mathbb N \}$ which is not compact.
This shows that if we drop the assumption that $\psi$ must be proper, $M'$ would be a submanifold of $\mathbb R^2$.