Does $\mathbb{R}^2$ Contain Uncountably Many Disjoint Copies of the Warsaw Circle?

The Warsaw Circle is defined as the closed topologist's sine curve, with an additional arc attached at its free end point and one of the end points of the critical line:

Since we don't have an uncountable collection of disjoint open sets in the plane - such as the bounded component of its complement - we'll have to have an uncountable sequence $W_r$ such that if $r < s$ then $W_r$ is contained in the bounded component of $W_s^c$. Trying to draw it, it's actually kind of hard for me to tell for sure if it's true or not; it LOOKS true, but jeeze I think showing it with analytic, coordinate representations of each one is gonna be a bear.

Anyone really feeling it?

To be honest, I don't know the answer for the closed topologist's sine curve itself, either! Probably both are possible or both are impossible; have never seen it proved for either space.

Yes, there exists such a collection. Any arc-like or circle-like planar continuum (all arc-like continua are planar) admits a mutually disjoint, uncountable collection of copies of itself in the plane. The proof is a bit long but elementary: https://link.springer.com/article/10.1007/BF01696773

Logan Hoehn has given an example of a tree-like continuum which is not arc-like and also admits uncountably many mutually disjoint copies in the plane. It's harder but still readable: https://uts.nipissingu.ca/loganh/CopiesOfX.pdf

A classification of continua which admit such embeddings is not known, and there isn't any generally accepted "main conjecture" on the problem. There's a broad feeling that if current tools are sufficient for such a classification then it will center around one of the various notions of 'span.'