Prove if there are 4 points in a unit circle then at least two are at distance less than or equal to $\sqrt2$ There is a unit circle and 4 points inside the circle. The problem is to prove at least two are at distance less than or equal to $\sqrt2$
 A: An idea: taking the circle $\,S^1:=\{(x,y)\in\Bbb R^2\;;\;x^2+y^2=1\}\,$ in order to use some analytic geometry/linear algebra in case of need, we can argue as follows:
If two points $\,w_i:=(x_i,y_i)\;,\;i=1,2\;$ , lay on the same quadrant (including the respective parts of both axis in the quadrant), then their maximal distance is $\,\sqrt2\;$ , because 
$$||w_1-w_2||^2=||w_1||^2+||w_2||^2-2\langle w_1\,,\,w_2\rangle$$
But
$$\min_{w_i\in S^1}|\langle w_1\,,\,w_2\rangle|=\min_{w_i\in S^1}||w_1||\,||w_2||\cos\theta=0\iff \theta=\frac\pi2\implies$$
$$||w_1-w_2||^2\le ||w_1||^2+||w_2||^2\le2$$
After the above one can see that the maximal possible distance between four points on the unit circle or within the unit disk is attained when they are on the intersection of the circle with the axis, i.e. on the points $\,(-1,0)\,,\,(1,0)\,,\,(0,-1)\,,\,(0,1)\,$ , and then the distance between any two consecutive (clock or anticlockwise) points is precisely $\,\sqrt2\,$ (between antipode points it is, of course, $\,2\,$ ...)
A: Suppose $O$ is the center of the circle. Now take $4$ points $A_1,A_2,A_3$ and $A_4$ on the circumference of the circle such that $OA_1$ passes through one of the $4$ points inside the circle and $\angle A_iOA_{i+1}=\pi /2$ for each $i$. Now consider the sector $A_1OA_2$; note that $|A_1O|=|A_2O|=1\leq\sqrt{2}$ and $|A_1A_2|=\sqrt{2}$. Hence any two points in the sector $A_1OA_2$ must have distance less than or equal to $\sqrt{2}$ and the same holds for all other sectors. Now since one point is on the boundary of two sectors, after you choose other three points at least two must be inside or on the boundary of the same sector.
A: here is a geometry way to prove the max distance is $\sqrt{2}$ is a quadrant area:

$A,B$ are the two points in area $FOG$, WLOG, let $AO \ge BO$,draw a circle with $r=AO$, cross $OB$ at $C$. 
look at $\angle CBA$, there are 2 cases:
case I: if  $\angle CBA< \dfrac{\pi}{2}$, in $\triangle ABO, \angle ABO > \dfrac{\pi}{2} \implies AB < AO \le 1 $, 
case II: if $\angle CBA \ge \dfrac{\pi}{2} $, in $\triangle ABC,\angle CBA$ is max, $\implies AC \ge AB$,(when $B=C, AC=BC$)  
it is trivial that $\triangle DOE \sim \triangle AOC ,DO\ge AO \implies DE \ge AC  \implies  AB \le DE  $
in $\widehat { FG}, \widehat { FG} \ge \widehat {DE} \implies FG \ge DE \implies AB \le FG=\sqrt{2}$
To summary above 2 cases, $AB_{max}=\sqrt{2}$ when and only when $A=F$ and $B=G$
The rest induction is same as DonAntonio's post. 
