A question on Möbius strip and Jordan curve If $A\subset \Bbb R^2$ then is the following statement true?

$\{(x,y)\in  {(A\times   A)/ \sim}\,\,\,|\,\, (x,y)\sim(y,x)\}\simeq$ Möbius strip $\iff A$ is a Jordan curve.

 A: It seems that the answer to this problem is affirmative. We can argue as follows.
Assume that $A\subset\mathbb R^2$ is a subspace whose symmetric square $A^2/_\sim$ is homeomorphic to the Mobius strip. In particular, $A^2/_\sim$ is a connected 2-manifold with a boundary. Using this fact it can be shown that $A$ is connected and locally path-connected. Since the product $Y\times I$ of the triod $Y$ and the interval $I$ does not embed into a 2-manifold, the space $A$ does not contain triods (i.e., subsets homeomorphic to the letter $Y$). Combining this fact with the local path-connectedness, we can prove that each subset $J$ of $A$, homeomorphic to an open inteval in the real, is open in $A$. Then $A$ contains an open dense subset, which is 1-manifold. Using the local path-connectedness and the absence of triods in $A$, it can be shown that $A$ is a 1-manifold with a boundary. So, $A$ is homeomorphic to a circle or to one of the intervals: $[0,1]$, $[0,1)$ or $(0,1)$. Since symmetric squares of  untervals are triangles (not homeomorphic to the Mobous strip), $A$ must be homeomorphic to the circle and hence $A$ is a Jordan curve in the plane.
