Let $\gamma: S^1 \mapsto \mathbb R^2$ be a simple closed curve (with positive orientation) in the plane that contains the origin in its interior. I would like to prove that it is homotopic to a small cocentric circle (in the punctured plane) with positive orientation, i.e, there is a continuous map $ [0,1]^2 \ni (s,t) \mapsto H(s,t) $ such that
$$ H(0,t) = \gamma (t), \quad H(1,t) = re^{2\pi it}, \quad \text{for some} \quad r >0. $$
So I consulted this post Homotopy of Jordan Curves?, I understand that if we find a homotopy between $\gamma$ and a bigger circle in the exterior of $\gamma$ and then use the standard homotopy between circles to finish the construction, but I hope to get a much simpler argument (if there is any!) in this thread. Any hints are highly appreciated!
By the Jordan curve theorem, $\text{ext}(\gamma)$ is connected and therefore $\overline{\text{ext}(\gamma)} = \mathbb{R}^2 \setminus \text{int}(\gamma)$ is connected. So then by 7 => 1 here https://en.wikipedia.org/wiki/Riemann_mapping_theorem, $\text{int}(\gamma)$ is simply connected. Then just make a perturbation argument to show that $\gamma \sim \gamma'$ where $\gamma'$ lies inside $\text{int}(\gamma).$ For example, if $\gamma$ is a smooth curve, then for each point $p$ on $\gamma$, just perturb $\gamma$ in the direction of the normal vector pointing towards the interior region. In general, this seems to follow from the proof of the Jordan curve theorem, namely, the homology version shows $H_1 = 0$. Alternately, via the Riemann mapping theorem, there is a map $\phi \colon \text{int}(\gamma) \to D$ that extends to a map $\overline{\text{int}(\gamma)} \to \overline{D}$. So given a homotopy $H$ in $D$, $\phi^{-1} \circ H \circ \phi$ is a homotopy in $ \text{int}(\gamma)$. Either way, the crux of the argument is that the Jordan curve theorem implies there are two connected components that are relatively "nice". This doesn't construct such a homotopy, but I imagine that would be difficult depending on the complexity of $\gamma$.