The problem is: given $A = (0,1) \subset \mathbb{R}$, find a set $B \subset \mathbb{R}^2$ and a function $g: A \longrightarrow B$ which is a continuous bijection but not a homeomorphism.
So I know that in order for $g$ not to be a homeomorphism, its inverse must not be continuous. I have seen a similar example of a function $f: (0,1] \longrightarrow S^1$ which not a homeomorphism since $f^{-1}$ is not continuous at $(1,0).$ However, if the domain of $f$ were instead $(0,1)$ it would be a homeomorphism, so I'm not sure how (or if) $S^1$ would work in the problem for the codomain of $g$.
I've looked at the answers to the other questions on Math.SE pertaining to examples of non-homeomorphic continuous bijections but none have seemed to help, e.g. the domain is usually more like $(0,1]$ than $(0,0)$.
Best Answer
You can do this with a figure eight space, e.g. map $(0,\tfrac12]$ to the circle centered at $(0,1)$ with radius $1$ and map $[\tfrac12,1)$ to the circle centered at $(0,-1)$ likewise with radius $1$ and in both cases with $\tfrac12$ mapping to the origin. Then "glue" the maps together to get one continuous bijection from $(0,1)$ to the figure eight. It won't have a continuous inverse because a figure eight minus one point (with one obvious exception) will be connected, but $(0,1)$ minus one point isn't connected.
Such a map is an example of an immersion.