Prove that squares split the plane into regions that have boundary equal to some Jordan curve.

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In Tao's proof of the Jordan curve theorem in the appendix of his 246A Notes 3 he covers a simple closed curve $\gamma$ with squares of small sidelength and claims that "the boundaries of these squares divide the complex plane into a finite number of polygonal regions". I provide his illustration: enter image description here I want to make that rigorous as follows: for a collection of square curves $\gamma_{\square, 1}, \ldots, \gamma_{\square, n}$, the complement of the union of their images is an open set; then each connected component of this open set has boundary equal to a disjoint union of simple closed curves (consisting of pieces of the original $\gamma_{\square,i}$ of course).

I think this is true for say circles as well, but I am having a very hard time proving the claim in both the case for squares and circles. Caveat: there are probably some issues with simplicity of the curves, but we can perturb the squares/circles a bit so we can basically ignore that issue.

EDIT: seems like this question has pretty much already been asked before: Jordan Curve Theorem, Professor Tao's proof.