as I am not an expert on differential topology, but am working with SDEs on submanifolds anyway, I have come across the following Proposition in Elton P. Hsu's Stochastic Analysis on Manifolds.
[Proposition 1.2.6, p. 21]
Let $M$ be a (noncompact) closed (equivalently properly embedded) submanifold of $\mathbb{R}^n$ without boundary, where $n = 2\text{dim }M+1$ (Whitney) and $\hat{M} = M \cup \{\partial_M \}$ its one-point compactification. A sequence of points $\{ x^n\}$ in $M$ converges to $\partial_M$ in $\hat{M}$ iff $\vert x \vert_{\mathbb{R}^n} \rightarrow \infty$.
$\textit{Proof(sketch)}:$ "$\Leftarrow$". Assume $x^n$ converges in the Euclidean metric on $\mathbb{R}^n$ to $\infty$. As $M$ is a closed subset of $\mathbb{R}^n$ every sequence in $M$ has a limit in $M$. In other words $\overline{M} \subset \mathbb{R}^n$. Now since $\vert x^n \vert_{\mathbb{R}^n} \rightarrow \infty$, at least one of $x^n$'s coordinates expressed in coordinates of $\mathbb{R}^n$ is unbounded. But $\infty \not\in \mathbb{R}^n$, so $x^n$ has to exit the subset $M$ eventually as $n \rightarrow \infty$, i.e. $x^n \rightarrow \partial_M$ in $\hat{M}$.
"$\Rightarrow$". Assume now the converse, then for any point $p \in M$ the distance between the sequence $x^n$ and $p$ becomes infinitely large as $n \rightarrow \infty$. But since $M$ is a closed subset of $\mathbb{R}^n$ every point on $M$ has finite distance to the origin. Hence, $\vert x^n - 0\vert_{\mathbb{R}^n} \rightarrow \infty$ as $n \rightarrow \infty$.
$\square$ ?
Now from a differential topologists view, is that an admissible proof, or have I missed important aspects?
Thank you.