This video discusses a topological proof of the following theorem:

Given a Jordan curve $C$, there exist four points on $C$ that are the vertices of a rectangle.

I'm not aware of any non-topological proof.

I'll sketch the proof. Let's begin with an equivalent restatement:

There exist distinct unordered pairs of points on $C$ of the same length and midpoint (namely, the rectangle's diagonals).

(If we wanted to find a square in $C$, the diagonals would need to be orthogonal, which isn't a property they have in common so much as a binary relation between them. This explains why, as the video notes, we haven't been able to adapt the techniques below to proving the always-a-square conjecture.)

Assume $C$ lives in the plane $z=0$, and consider the function $f:C\times C\to \mathbb{R}^3$ defined by $f(a,\,b):=\left(\frac{a_x+b_x}{2},\,\frac{a_y+b_y}{2},\,|b-a|\right)^T$ so $f(a,\,a)=(a_x,\,a_y,\,0)^T=a\in C$, and $S:=\mathrm{rang} f$ is a 2D surface in $\mathbb{R}^2\times [0,\,\infty)$ with $\partial S = C$. The claimed theorem can now be equivalently restated again:

The surface $S$ is self-intersecting, just like the Möbius strip.

In fact, we'll prove these two surfaces are topologically equivalent.

Endow $C$ with an affine parameter $\lambda$ running from $0$ to $1$, both extrema applying to the same point $p\in C$, with all other $q\in C$ having a unique value of $\lambda$. We can identify $(a,\,b)\in C^2$ with points on a square, with parallel edges joined as in "Space Invaders" topology. By contrast, an unordered pair $\{ a,\,b\}\subset C$ is identified with a point in the shape obtained as follows:

- Fold the square along $y=x$ to form a triangle, so $\{ a,\,b\}$ isn't distinguished from $\{ b,\,a\}$.
- We can't join the edges as before in an orientation-preserving way yet, so first cut the square along its other diagonal, $y=1-x$.
- Rotate the triangles so the orientation-preserving joins can be made after all. But this introduces a half-twist; unsurprisingly, the result is the Möbius strip. (Feel free to try it with paper, scissors & glue.)