Is it possible to have a tessellation of the plane with no corners?
Here's what I mean precisely: Is it possible to tile the plane with tiles, each of which is the region enclosed by a smooth simple closed curve? Also assume that only finitely many tiles in any bounded region.
I don't think so, because when two of these smooth simple closed curves separate from each other, they should leave some sort of cusp that can't be filled by a smooth simple closed curve, but I'm not sure how to make this rigorous.
Idea:
Let $T$ be a tile and let $C$ be the smooth simple closed curve that bounds it. Traversing $C$ we pass a finite number of other tiles $T_1,\ldots,T_n$, because any bounded region contains only finitely many tiles. Note that $n=1$ is impossible as then $T_1$ is not enclosed by a smooth simple curve. Then $n\geq2$, and so $T$, $T_1$ and $T_2$ meet at a point $p$. At this point $p$ at least one of the three boundaries is not smooth by handwaving.
As suggested in the comments by Thomas Browning: As the boundaries of the tiles are smooth simple closed curves, their tangent vectors are well-defined at every point. Then for every pair of tiles meeting at $p$, traversing their boundaries in clockwise direction yields tangent vectors pointing in opposite directions. Repeating this three times then shows that the tangent vectors at $p$ are opposite to themselves, a contradiction.
Original answer:
It depends on what you mean by a tessellation: If you consider a covering of the plane by concentric annuli a tessellation, then there's an example for you.