Punctured plane is not simply connected Adapt the following definition of "simply connected space" (taken from Wikipedia):

A space $X$ is simply connected if it's path connected and for any continuous map $f:S^1\rightarrow X$ can be extended to a continuous map $F:D^2\rightarrow X$ such that $F\mid_{S^1}=f$.

Then it's well known (and obvious?) that the punctured plane $\Bbb R^2\setminus\{(0,0)\}$ with standard topology is not simply connected, because unit circle is not contractible (i.e. $f$ mapping $S^1$ to unit circle in plane cannot be extended to $F$ as in definition above).
However, I have never seen an easy proof of this fact. My question is the following:

Is there a proof that punctured plane is not simply connected which doesn't use any results beyond Jordan curve theorem or Jordan–Schoenflies theorem?

If not, do you know of some proof of this fact which could nevertheless be considered elementary (in a way the proof of JCT using Brouwer fixed-point theorem is considered elementary)?
Thanks in advance.
 A: Let $X$ be the complement of the origin in $\mathbf{R}^{2}$. If there existed a continuous map $F:D \to X$ extending the inclusion $f:S^{1} \to X$, Green's theorem applied to the smooth $1$-form
$$
\omega = \frac{-y\, dx + x\, dy}{x^{2} + y^{2}}
$$
would give
$$
0 = \iint_{F(D)} d\omega
  = \int_{\partial F(D)} \omega
  = \int_{f(S^{1})} \omega
  = 2\pi.
$$
(Spivak's Calculus on Manifolds develops the theory of integration of smooth differential forms over continuous chains, and his proof of Green's theorem amounts to little more than the one-variable FTC, so I think the entire proof satisfies the spirit of "elementary".)
A: Spoiler alert: The following is actually a result of homotopy theory in disguise.

Let suppose that $X=\mathbb R^2 - (0,0)$ is simply connected. Then the antipodal map
$$\begin{aligned}f:S^1&\rightarrow X,\\
x&\mapsto -x
\end{aligned}$$ can be extended to a map
$$\tilde f:D^2\to X.$$
Now, consider the map
$$g:D^2\stackrel{\tilde f}\to X\to S^1\hookrightarrow D^2,$$
where the middle map is $x\mapsto \frac{x}{\|x\|}$. Obviously the image of $g$ is $S^1$. Thus the only possible fixed points of $g$ are on $S^1$, whose images are their antipodals. Therefore, $g$ does not have any fixed points which contradicts Brouwer's fixed point theorem.
