It is a corollary of the Riemann mapping theorem that every path-connected simply-connected open subset of $\mathbb R^2$ is homeomorphic to $\mathbb R^2$ (there are more elementary proofs as well https://mathoverflow.net/questions/66048/riemann-mapping-theorem-for-homeomorphisms).
This is of course not true for closed subsets $C \subseteq\mathbb R^2$; in fact the invariance of domain theorem tells us that no proper closed subset $C \subsetneq \mathbb R^2$ can be homeomorphic to $\mathbb R^2$.
I am interested however in topological invariants that can tell spaces apart, like path-connectedness, simply-connectedness, compactness, etc.; using the invariants I just mentioned, we see that if a closed subset $C \subseteq \mathbb R^2$ is homeomorphic to $\mathbb R^2$, then it must be path-connected, simply-connected, and not compact (i.e. $\iff$ unbounded, by Heine-Borel in $\mathbb R^2$).
The "next hardest example" not covered by the above is e.g. $C:=$ a closed half plane. We can rule that out by removing any point $b$ on the boundary of the closed half plane $C$; then $C \setminus \{b\}$ is still simply-connected, while $\mathbb R^2$ removed any point is not simply-connected.
So, the "next hardest example" would be a proper closed set $C \subsetneq \mathbb R^2$ that is path-connected, simply-connected, but removing ANY point of $C$ results in a still path-connected, but no longer simply-connected space. I will call such a set "fragilely"/“minimally” simply-connected (because it is simply-connected, until you remove ANY point) and "robustly" path-connected (since it is path-connected, and remains so after removing ANY point).
Question: does a "fragilely" simply-connected, "robustly" path-connected closed proper subset $C \subsetneq \mathbb R^2$ exist?
Best Answer
No such set exists. More strongly, any fragilely simply connected robustly path-connected subset of $\mathbb{R}^2$ must be open. To prove this, suppose $C\subset\mathbb{R}^2$ is fragilely simply connected and robustly path-connected but not open. Let $p\in C$ be a boundary point.
Now we invoke a hard theorem (see here): if $C\setminus \{p\}$ is path-connected but not simply connected, then there is a simple closed curve $\gamma$ in $C\setminus\{p\}$ and a point $q\in\mathbb{R}^2\setminus (C\setminus\{p\})$ in the interior of $\gamma$. Since $C$ is simply connected, this point $q$ cannot be in $\mathbb{R}^2\setminus C$, so it must be $p$. Since $p$ is a boundary point of $C$, it can be approximated by points not in $C$. Since the interior of $\gamma$ is open, this means there is a point $p'\in\mathbb{R}^2\setminus C$ that is also in the interior of $\gamma$. But this then implies that $\gamma$ is not nullhomotopic in $C$, which is a contradiction.