Circles and pigeonhole

Arcs. Divide circle into $3$ equal arcs of $2\pi/3$.

Pigeonhole. $4$ points, $3$ arcs: some arc has $\ge2$ points.

Distance. Max chord length in an arc of $2\pi/3$ is $2\sin(\pi/3)=\sqrt3>\sqrt2$. So this doesn't give the claimed bound.

Corrected. Divide into $4$ arcs of $\pi/2$, giving chord length $\sqrt2$.

Conclusion. Among $5$ points on a circle, two are within chord $\sqrt2$ (chord of $\pi/2$ arc). With $4$ points and $4$ arcs, no pigeonhole forces $2$ in one arc.

Conclusion (corrected). $5$ points guarantee two within $\sqrt2$.