The letters of the word "ART" 
Edit: According to the Gelfand duality between topological spaces and  commutative $C^{*}$algebras, I add some new tags. So the question is that what is the structure of  $ Ext (A,A)$  where $A$ is $C_{0}(\mathbb{R})$. One can think to stabilization of this question, that is $A=C_{0}(\mathbb{R})\otimes \mathcal{K} $  where $\mathcal{K}$ is  the  algebra  of  compact operators.

The  following was the  first version of my question:
Are there only a finite number of connected topological spaces $X$ (up to homeomorphism) with the property that $X$ has an open subset $U$ such that $U$ and $X-U$ are homeomorphic to $\mathbb{R}$? I know three examples as I wrote in the title of this question. (We delete the end critical points from each letter.) Among capital alphabet, there are no other topological type with the above property.
Is it true that any space $X$ with this property can be embedded in $\mathbb{R}^{2}$?
 A: Yes there are infinitely many, by a version of Mike Jury's idea. In fact there are uncountably many embeddable in $\mathbb R^2$. Take the union of two real curves:
the open one $y= x^{-1}\sin(1/x)$
the closed one $x= f(y)$ for $f$ a function that is $0$ on some closed set $S$ and negative elsewhere. 
The closure of the open curve is exactly $S$, so we can identify $S$ from the topology of the space. Clearly there are uncountably many closed subsets of $\mathbb R$ up to homeomorphism of $\mathbb R$ (e.g. encode a real number as a sequence of closed intervals, isolated points, and Cantor sets.) So there are uncountably many spaces.
A: There is a nonmetrizable space $X$ with that property so the embedding property fails.
I don't know if one could build more elaborate examples along these lines to find infinitely many such spaces.
The answer to your question may be different if you require $X$ to be metrizable or Hausdorff or otherwise nice.
Let $A$ and $B$ denote two copies of $\mathbb R$ with the usual topology and let $X$ be the disjoint union of $A$ and $B$.
Define a topology on $X$ so that $\alpha\cup\beta$ with $\alpha\subset A$ and $\beta\subset B$ is open iff


*

*$\alpha$ and $\beta$ are open (in the usual real topology) and

*$\beta=B$ or $\alpha=\emptyset$.


This is indeed a topology.
The union of any two disjoint nonempty open sets does not meet $A$, so $X$ is connected.
The topology is not Hausdorff and thus not metrizable; two distinct points in $A$ cannot have disjoint neighborhoods.
Now $B\subset X$ is open and its complement is $A$, and both of these subspaces are homeomorphic to the real line.
