I'm a huge fan of hyperbolic geometry. I find the Poincaré disk pretty impressive because it represents an infinite space within a finite space.
Is there a way to show the infinite Euclidean 2D space (aka an Euclidean plane) in a finite way? Is it feasible or just unthinkable?
$\newcommand{\Reals}{\mathbf{R}}$If $\phi:\Reals^{2} \to D$ is a bijection to a bounded set, then $D$ can be viewed as a model of the Cartesian/Euclidean plane. With no further constraints, however, such a model is unlikely to satisfy: Lines map to (effectively) arbitrary curves, rigid motions look "far from rigid" in $D$, and so forth.
Here are a couple of "relatively nice" representations of the Euclidean plane in a bounded spatial or planar region:
Stereographic projection from $(0, 0, 1)$ (below, left) $$ \phi(u, v) = (x, y, z) = \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1},\qquad \phi^{-1}(x, y, z) = (u, v) = \frac{(x, y)}{1 - z}, $$ maps the plane to the complement of the north pole $N = (0, 0, 1)$ in the unit sphere $S^{2} \subset \Reals^{3}$.
These maps are conformal (angle-preserving), Euclidean lines map to circles through $N$, and Euclidean rigid motions look not unlike hyperbolic rigid motions acting on the Poincaré disk.
Central/gnomonic projection from $(0, 0, 1)$ (below, right) $$ \phi(u, v) = (x, y, z) = \frac{(u, v, -1 + \sqrt{u^{2} + v^{2} + 1})}{\sqrt{u^{2} + v^{2} + 1}},\qquad \phi^{-1}(x, y, z) = (u, v) = \frac{(x, y)}{1 - z} $$ maps the plane to the "southern hemisphere" $x^{2} + y^{2} + (z - 1)^{2} = 1$, $0 \leq z < 1$. (If desired, map the open hemisphere to the open unit disk, either by vertical projection, or by stereographic projection from $(0, 0, 2)$.)
Every Euclidean line determines a unique plane through the center of the sphere, and therefore maps to half a great circle. Again, Euclidean rigid motions look not unlike hyperbolic rigid motions acting on a disk model of the hyperbolic plane.