Using a cut-point to break a homeomorphism

Question

How would I go about using the fact that having a cut-point is a topological property to show that $ \mathbb{R} ,\mathscr{U} \rightarrow \mathbb{R},\mathscr{C} $ is not a homeomorphism where $ \mathscr{U} $ is the usual topology on $ \mathbb{R} $, given by $ \mathscr{U} = \{U | x \in U \rightarrow x \in (a,b) \subseteq U\} $ and $ \mathscr{C} = \{\emptyset \} \cup \{(a,\infty) | a \in \mathbb{R} \} \cup \{\mathbb{R} \} $ is the open half-line topology.

Answer 1

Why use cut points? Clearly R = usual R is Hausdorff while H = open
half line R is not Hausdorff. Hence R and H aren't homeomorphic.

If you must use cutpoints, assume f:H -> R is homeomorphism.
R' = R - {f(x)} is disconnected; H' = H - {x} is connected.
g = f restricted to H' is continuous surjection onto R'.