For this question I will consider $\mathbb{R}^2 = \mathbb{C}$ only to use the Riemann Sphere.
The Jordan Curve Theorem says a jordan curve divides divides the space into a interior region, bounded by the curve and the exterior, unbounded.
The term bounded means the set is limited (exists a circle which contains all the bounded set).
Clearly a circle (a jordan curve) $x^2+y^2=1$ divides the plane into two regions $S_1$ and $S_2$, and so does the line $y - x= 0$, but both regions are unbounded.
Although the idea of "line divides a plane in two" seems intuitive, I cannot say for sure any curve that goes to infinite in both ways divides a plane in two regions (clearly unbounded).
So, interested of finding a condition to decide if a "infinite" curve divides the plane in two:
Question: If I pick a curve and project it in Riemann Sphere, and the resultant curve is a jordan curve, does it means the original curve divides a plane in two? Is the versa true? Are there counter-examples?
For example, the hyperbola $y^2-x^2=1$ is mapped into Riemann Sphere like an $8$-ish shape with the intersection at $\infty$ and therefore this hyperbola divides the plane in three, not in two. A parabola would divide the plane into two regions.
The Jordan curve theorem for the plane implies a version of the theorem for the 2-sphere: if $\gamma : S^1 \mapsto S^2$ is continuous and injective with image $\cal I$, then $S^2 \setminus {\cal I}$ is homeomorphic to two open discs. To prove this pick $x \in S^2 \setminus {\cal I}$, take it as the point at infinity and use the resulting stereographic projection to translate the problem into a problem about a continuous injection $S^1 \to \Bbb{R}^2$. Then apply the Jordan curve theorem for the plane and pull your conclusions back along the stereographic projection.
Minor issue: how do we know $S^2 \setminus {\cal I}$ is non-empty? Answer: because $S^1$ and $S^2$ are compact, $\gamma$ maps closed sets to closed sets and hence, if it were surjective, it would be a homeomorphism; but $S^2$ and $S^1$ are not homeomorphic, because removal of two distinct points disconnects $S^1$ but not $S^2$.