Is simple connectedness of subsets of $\mathbb{R}^2$ detected by winding numbers?

Question

If $X\subset\mathbb{R}^2$ is a connected open subset which is not simply connected, then there exists some point $p\in\mathbb{R}^2\setminus X$ such that the inclusion map $X\to\mathbb{R}^2\setminus \{p\}$ induces a nontrivial map on $\pi_1$. In other words, if a connected open subset of the plane is not simply connected, this is always because it contains a loop which winds around some point in its complement. More strongly, in fact, there is some point $p\in\mathbb{R}^2\setminus X$ such that $X$ contains a simple closed curve that has $p$ in its interior. (See here for one way to prove this.)

My question is whether this is still true for arbitrary (not necessarily open) subsets of $\mathbb{R}^2$. More precisely:

Suppose $X\subset\mathbb{R}^2$ is path-connected but not simply connected. Must there exist some point $p\in\mathbb{R}^2\setminus X$ such that the inclusion map $X\to\mathbb{R}^2\setminus \{p\}$ induces a nontrivial map on $\pi_1$? More strongly, must there exist such $p$ such that $X$ contains a simple closed curve that has $p$ in its interior, or at least a curve with winding number $1$ around $p$ (so the map on $\pi_1$ is surjective)?

I suspect the answer is yes, since I've heard of various results saying that arbitrary subsets of $\mathbb{R}^2$ are homotopically "tame" in various ways (for instance, unlike in higher dimensions, they cannot have nontrivial $H_k$ for $k\geq 2$). The closest result to my question that I know of is the result from this answer that if $X\subseteq \mathbb{R}^2$ then any nontrivial element in $\pi_1(X)$ remains nontrivial in $\pi_1(\mathbb{R}^2\setminus F)$ for some finite set $F\subseteq \mathbb{R}^2\setminus X$. This is very close to my question: it is weaker since it involves a finite set $F$ rather than a single point, but stronger since it applies to an arbitrary nontrivial element of $\pi_1(X)$ rather than just saying there exists an element of $\pi_1(X)$ that remains nontrivial. I would guess you can prove an affirmative answer to my question either from that result or using ideas similar to its proof, but I don't see an obvious way to do so. (Note that the finite set $F$ in that result cannot be reduced to a single point, since your element of $\pi_1(X)$ could be a commutator of loops around different points.)

(Alternatively, by the argument I linked above in the case of open subsets, it would suffice to show that there exists a nontrivial loop in $X$ with only finitely many self-intersections, since then you can use that to reduce to a nontrivial simple loop and then invoke the Jordan curve theorem.)

Answer 1

The strong version of my question has been answered affirmatively by Jeremy Brazas on MathOverflow:

All the ingredients to prove what you want (and more) can be found in

H. Fischer, A. Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655--1676.

Proof Sketch. Suppose $X\subseteq \mathbb{R}^2$ is path connected and $\alpha:S^1\to X $ is not null-homotopic. Let $Y=\alpha(S^1)\cup\bigcup\{U\mid U\text{ is a bounded path component of }\mathbb{R}^2\backslash \alpha(S^1)\text{ with }U\subseteq X\}$. Fischer and Zastrow show $Y$ is a Peano continuum homotopy equivalent to a “Sierpinski-like” space, i.e. a copy of the Sierpinski Carpet with some, possibly empty, set of (originally deleted) squares filled back in. Since all we want to do is find some simple closed curve that doesn’t contract, pick any bounded path component $C$ of $\mathbb{R}^2\backslash Y$ and a point $z\in C\backslash X$. While $C$ is homeomorphic to an open disk, its boundary need not be a simple closed curve. However, you basically have a planar Peano continuum $Y$ surrounding a “hole” at $z$. Since $Y\subseteq X$ and $z\notin X$, any simple closed curve in $Y$ with winding number $1$ around $z$ will do the trick. If you want to create one explicitly, use some planar geometry to cover $Y$ with enough arc-wise connected open (in $Y$) sets to create a finite loop of neighborhoods in $Y$ with winding number $1$ around $z$ and build your curve piece-wise.