In their book Differential Topology, the authors Guillemin and Pollack give the following counterexample to the claim that the image of an injective immersion $f: X \to Y$ must necessarily be a submanifold of $Y$:
Let $g : \Bbb R \to S^1$ be the local diffeomorphism $g(t) = (\cos 2\pi t, \sin 2\pi t)$. Define $G : \Bbb R^2 \to S^1 \times S^1$ by $G(x, y) = (g(x), g(y))$. $G$ is a local diffeomorphism of the plane onto the torus. In fact, looking at $G$ on the fundamental unit square, one may consider it to be a construction of the torus by gluing opposite sides of the square together. Now define a map of $\Bbb R$ into the torus by restricting $G$ to a straight line through the origin in $\Bbb R^2$ with irrational slope. Since G is a local diffeomorphism, this is an immersion that wraps $ \Bbb R$ around the torus. Moreover, the irrationality of the slope implies that the immersion is one-to-one and that its image is a dense subset of the torus!
On explaining what goes wrong here they continue:
Notice that the injective immersion behaves strangely because it maps too many points "near infinity" in $\Bbb R$ into small regions of the image. Perhaps prohibiting this behavior will sufficiently tame immersions. The general topological analog of points "near infinity" is the exterior of a compact subset in a given space, the compact set being thought of as very large.
Now, my interpretation of "mapping too many points "near infinity" into small regions of the image" is if we take any $\epsilon$ ball $B_{\epsilon}$ such that $B_{\epsilon}\cap S^1 \times S^1 \neq \phi$ and any $R>0$, we can find infinitely many points $x>R$ such that $f(x)\in B_{\epsilon}\cap S^1 \times S^1$. Is my interpretation correct? And if so, how does that cause a problem here?
Why does the topological analog of points "near infinity" require the set to be compact? I don't see why it cannot be thought of as any large set. What's so special about compactness here?