Homeomorphism between plane with different topologies How would you show that spaces $(\mathbb{R^2},\cal{T}_r)$ and $(\mathbb{R}^2,\cal{T}_b)$, where $\cal{T}_r$ is a topology generated by jungle river metric (here) and equivalently $\cal{T}_b$ is generated by the British Rail metric (3.15 second one), are not homeomorphic? 
I tried to think of any elementary topological property, with which I can characterize only one of those spaces, but they seem to be so alike... Yet not homeomorphic. Why?
 A: In the railway system, there exists a (unique) point such that its complement is the disjoint union of infinitely many connected open sets.
No such points exists in the jungle. In short: The "main stream" is like $\mathbb R$ with standard metric, hence has at most two connected components after removing a point. But the corresponding open sets must reach into each of the vertical "arms", which look like $\mathbb R$ as well. 
To see this more explicitly: For any point $(x_0,y_0)$ the subsets $\{\,(x,y)\mid x<x_0\,\}$, $\{\,(x,y)\mid x>x_0\,\}$, $\{\,(x_0,y)\mid y<y_0\,\}$,  $\{\,(x_0,y)\mid y>y_0\,\}$ are even path connected (along the obvious polyline paths that the metric suggests), hence $\mathbb R\setminus\{(x_0,y_0)\}$ has at most four connected components.
A: Removing the origin from the river metric gives us 4 path-components (the positive $y$-axis, the negative $y$-axis, and the sets $\{(x,y): x > 0\}$ and $\{(x,y): x < 0\}$.
If we remove a point from the railway metric, then either we get uncountably many path components (if we remove the origin), or 2 (for all other points). 
