Does a Jordan curve partition sufficiently small balls centered on it into an interior and exterior component

Question

I am trying to prove the following, which seems to be intuitively true:

For every point on a Jordan curve, sufficiently small disks centered around the point are partitioned by the curve into two connected components, one lying in the interior of the curve, and one lying in the exterior of the curve, where we don't include the curve in each component.

Any help would be appreciated. It would be fine to prove it for just polygonal paths, so the points on the curve would lie in the interior of edges or on vertices.

EDIT: I think I managed to come up with a proof in the case of a polygon. Fix a polygon and a point $P$ on its boundary. There are finitely many vertices so finding a disk around $P$ that does not contain any vertices (except possibly $P$) is not an issue. There are also finitely many edges on which $P$ does not already lie ($P$ either lies on one edge or $P$ is a vertex from which two edges emanate); so we pick a radius of the disk that is smaller than the shortest distances from $P$ to each of those other edges. So now we know that a disk exists which is cut into two sectors that do not contain any boundary points (we have removed the edge/s on which $P$ lies as well). So each sector can contain only interior and exterior points of the polygon. But sectors are connected, and any path connecting an interior point to an exterior point must run through a boundary point, which we know to not exist in a sector. So each sector has only interior points or only exterior points. Finally all disks around boundary points contain at least one of each of an exterior point and interior point. So one sector has only interior points and the other sector has only exterior points.

Please let me know what you think, and if you have any ideas for the general case.

Answer 1

HINT:

consider the graph of the function $x \sin\frac{1}{x}$ around $0$ ( a part of a Jordan curve ) .

Note: I took the suggestion of @Paul Frost: to change to the damper $x$ rather than the very weak one $\frac{1}{\log 1/|x|}$, or the initial $\sqrt{x}$.

$\bf{Added:}$ Note that by Jordan-Schoenflies' Theorem every Jordan curve in $\mathbb{R}^2$ is equivalent under a homeomorphism to a circle. Since for the circle we can find the corresponding disks around any point, it follows that for any point on a Jordan curve there exists a fundamental family of neigborhoods homemorphic to disks that are cut into two parts by the curve. It is just that they may not be actual disks in some cases.