I am interested in giving an example for a closed, connected submanifold of a connected manifold which does not carry the subspace (or relative) topology.
With some help from Boothby (III.4 p.71-72), I believe an example could be the following:
Let $F : \mathbb{R} \to \mathbb{R}^2$ be given by
$$F(t) = \left( 2 \cos\left( t - \frac{1}{2} \pi \right), \sin 2 \left(t - \frac{1}{2} \pi \right) \right) {;}$$
the image is a figure eight. Now, let $g(t)$ be a monotone increasing smooth function on $- \infty < t < \infty$ such that $g(0) = \pi$, $\lim\limits_{t \to - \infty} g(t) = 0$ and $\lim\limits_{t \to \infty} g(t) = 2 \pi$ (e.g., we may use $g(t) = \pi + 2 \arctan{t}$). Then define $G : \mathbb{R} \to \mathbb{R}^2$ by composition of $g(t)$ with $F(t)$:
$$G(t) = F(g(t)) = \left( 2 \cos \left( g(t) - \frac{\pi}{2} \right), \sin 2 \left( g(t) - \frac{\pi}{2} \right) \right) {;}$$
note that $G(t)$ is an injective immersion. By definition, it follows that $G(\mathbb{R}) = \tilde{N}$ is a submanifold (or immersed submanifold) of $\mathbb{R}^2$.
From this point, it gets tricky and I am unsure of how to show whether or not $\tilde{N}$ is closed. Boothby notes that $\mathbb{R}$ is not homeomorphic (under $G$) to $\tilde{N}$ considered as a subspace of $\mathbb{R}^2$. It seems reasonable that $\tilde{N}$ is closed, if $\mathbb{R}^2 - \tilde{N}$ is open, though I would like to be able to justify this rigorously.
Is this an example of what I am looking for? If not, could someone point me in the right direction? Any hints or suggestions would be greatly appreciated.
Edit: The image of a connected space under a continuous map is connected, and thus $\tilde{N}$ is connected.
A more direct injective immersion that gives a figure 8 is the map $(-\pi/2,3\pi/2) \to \mathbb R^2$ given by $t \mapsto (\sin 2t, \cos t$). It's an immersion because the derivative never vanishes. By definition, the image of this map is an immersed submanifold. Moreover, the image is closed as a subset of the plane. One way to see this is to note that this image is exactly the locus of points $(x,y)$ such that $x^2 = 4y^2 (1-y^2)$. Therefore, it is the fiber of $0$ under the continuous mapping $(x,y) \mapsto 4y^2 (1-y^2) - x^2$, whence it is closed.
[This is Example 7.2 in Lee's "Introduction to Smooth Manifolds".]