So, I know that this isn't an embedded submanifold, since with the subspace topology inherited from $\mathbb{R}^2$, it is not locally Euclidean at $0$. Further, I also know that if I can parametrize it nicely, I will be able to say that it is the image of an injective smooth immersion and thus an immersed submanifold, but I can't seem to do that and don't know how to prove that it is not one.
[Math] Is $x^2 = y ^2$ an immersed submanifold of $\mathbb{R}^2$
differential-geometrydifferential-topologysmooth-manifolds
Related Solutions
The key is to find a set that is the image of two different injective immersions that induce different topologies on it. Here's a hint:
$\qquad$ 
EDIT: For an example where the two submanifolds are not homeomorphic, try this:
In one case, the submanifold has two connected components, neither of which is compact. In the other, there are three components, one of which is compact.
Immersed submanifold with the subspace topology is not an embedded submanifold. In fact, when we consider the underlying set of an immersed submanifold with the subspace topology, the resulting space need not be a manifold at all. For example, with the subspace topology, the figure-eight is not a manifold since you cannot find an open set in $\mathbb{R}^2$ such that its intersection with the figure-eight gives an open set for some coordinate chart including $0$.
For the second statement, note that in the subspace topology, you cannot find an open neighborhood for $0$ in the figure-eight. Try to think this for the irrational line in torus example. Try to find an open set(in the subspace topology) for any point in the irrational line. (This shows how different two topologies are)
Immersed submanifold topology is inherited from the immersion map $f :S \to M$. That is, $f(U) \subset f(S)$ is open iff $U \subset S$ is open. This topology by definition gives us the homeomorphism $f(S) \cong S$.
One should think immersed submanifold is the same(topologically) as domain of the immersion when s/he uses immersed submanifold topology and one should consider immersed submanifold more carefully when s/he uses submanifold topology because as observed above in the subspace topology image may or may not be a manifold. On the other hand, embedded submanifolds are always homeomorphic to domain of embedding in the subspace topology.
Best Answer
If "immersed submanifold" means "the image of an immersion, a map whose differential is injective everywhere", then the crossed lines are the image of two parallel lines under a simple map.
More explicitly, let's define a map from $X=\{(x,y)\in\mathbb{R}^2|y^2=1\}$ to $Y=\{(x,y)\in\mathbb{R}^2|x^2=y^2\}$ via $f(x,y) = (x,\text{sign}(y)x).$ It maps two parallel lines to your two crossed lines. Since $\text{sign}(y)$ is constant on the connected components, its non-differentiability at $y=0$ doesn't come up. So $f$ is differentiable, and is an immersion.
However as Jack Lee points out in the comments, the more usual definition of "immersed submanifold" is a subset with its own topology and smooth structure (not necessarily subspace) such that the inclusion is an immersion.
We can realize $Y$ as an immersed submanifold in this sense as well. Let's define $f,g,h\colon \mathbb{R}\to\mathbb{R}^2$ by $f(x)=(x,x)$, and $g(x) = (-e^x,e^x)$ and $h(x)=(e^x,-e^x).$ So $f$ parametrizes the line through the origin and quadrants III and I, $g$ parametrizes the ray through quadrant II (approaching but never reaching the origin), and $h$ parametrizes the ray through quadrant IV (also approaching but never reaching the origin). These functions define a topology on their image which is not the subspace topology. Consider for example that the space with its immersed topology has three connected components, while it has only one in the subspace topology. They give the set $Y=\{(x,y)\in\mathbb{R}^2|x^2=y^2\}$ the structure of an immersed submanifold of $\mathbb{R}^2$.