An injective nonpolygonal arc drawing is excluded by the page's finite polygonal plane-graph convention

The first coordinate of $\gamma(t)$ is $t$, so $\gamma$ is injective. For $t>0$ it is continuous, and $|t\sin(1/t)|\le t$ shows continuity at $0$. Thus [F1] makes it a continuous arc with distinct endpoints. It is not polygonal: the second coordinate vanishes at infinitely many points $t=1/(k\pi)$ accumulating at $0$ and changes sign between them, whereas a finite union of line segments either has only finitely many such crossings of the horizontal axis or contains a nontrivial segment of that axis. The latter is impossible here because $t\sin(1/t)$ is not identically zero on any interval.

Use the image of $\gamma$ as the drawing of the sole edge of a two-vertex abstract graph. The drawing is injective but its edge is not a polygonal arc in the sense of Polygonal arcs and polygons as non-self-intersecting finite unions of line segments in $\mathbb R^2$, so [L1] excludes this particular drawing from the page's class of plane graphs. This does not make the abstract one-edge graph nonplanar: drawing its edge as a straight segment gives a plane embedding.