I am trying to prove the following claim:
If $B$ is a simply connected subset of $\mathbb{R}^2$ and $b \in B$, then any loop in $B\setminus\{b\}$ which is not path-homotopic to a constant map is also not path-homotopic to a constant map when viewed as a loop in $\mathbb{R}^2\setminus\{b\}$.
Below I refer to this claim as (*). It is the missing link in my proof of another statement. Here is an outline of my approach to that original problem:
Show that any subset of $\mathbb{R}^2$ which is homeomorphic to $\mathbb{R}^2$ is open. [I am aware that this is a consequence of the invariance of domain, which requires advanced techniques for the general proof. I am hoping that this special case, 2- rather than n-dimensional and with homeomorphism not to an arbitrary open subset, but to all of $\mathbb{R}^2$ might be possible with more basic topology.]
Proof which requires (*): If $B$ is the subset and $b$ is any point in it, then $B \approx \mathbb{R}^2$ is simply-connected, while $B\setminus\{b\} \approx \mathbb{R}^2\setminus\{0\}$ is not. This means there is a loop $f$ in $B\setminus\{b\}$ which is not path-homotopic to a constant map in $B\setminus\{b\}$, but is in $B$. Since the image of $f$ is compact, there is a shortest distance from $b$ to it, and I want to prove that the open disc around $b$ with radius equal to that distance is included in $B$.
Suppose that a point $x$ in that open disc is not in $B$. Then $B \subseteq \mathbb{R}^2\setminus\{x\}$ and therefore $f$ is path-homotopic to a constant map in $\mathbb{R}^2\setminus\{x\}$ (because it is such in $B$).
But it would follow from (*) that since $f$ is not path-homotopic to a constant map in $B\setminus\{b\}$, it is also NOT path-homotopic to a constant map in $\mathbb{R}^2\setminus\{b\}$. Since $b$ and $x$ are both inside that disc, this would be a contradiction, because we can define strong deformation retractions of both $\mathbb{R}^2\setminus\{b\}$ and $\mathbb{R}^2\setminus\{x\}$, each onto the boundary of that disc, which agree outside of it; that is they each send $f$ to the exact same loop in the exact same retract space.