No continuous embedding of the Klein surface into $\mathbb{R}^3$. This is a standard result from algebraic topology. Using Alexander duality to compute homology, we produce a contradiction.

Given a Jordan curve $J$, the set of unordered and unequal pairs of points in $J$, denoted $S$, is an open Möbius band. Because every unequal unordered pair of points in the circle determines a unique point in $\mathbb{R}P^2$: Take the two tangent lines to the circle at these points and intersect them.

If tangent lines are parallel, than it indicates the infinite point of $\mathbb{R}P^2$:

Therefore, $S$ are the points of the complement of the closed unit disk in the projective plane, hence an open Möbius band.

Consider a map $\phi : S \to \mathbb{R}^3$ defined by

Geometrically, $\phi$ maps the ordered pair to a point encoding the midpoint of the segment $\overline{ab}$ (2D) and the length of the segment (1D).

If $\phi(a_1, b_1) = \phi(a_2, b_2)$, then these four points form an inscribed rectangle. Hence we just need to prove that the map $\phi$ is not injective!

Consider $K = \phi(S) \cup \phi(\partial S) \cup \rho(\phi(S))$ where $\rho$ reflects the shape in the XY-plane. This is a Klein surface by gluing two Möbius strips at their boundary.

If $\phi$ is injective, then $K$ can be embedded into $\mathbb{R}^3$, which contradicts the theorem that nonexistence of an embedded Klein surface into $\mathbb{R}^3$.