$2$ out of $4$ points, each of distance at most $1$ apart, are at most $1/\sqrt2$ apart Given $4$ points in the plane such that any pair of them are a distance of at most $1$ apart, show that some pair of them must be of distance at most $1/\sqrt{2}$  apart.
I figured the solution might involve a pigeonhole argument, but as my geometry is rusty I'm having trouble turning the "distance of at most $1$" condition into a condition that bounds all $4$ points in a region that can then be subdivided into $3$ regions in which every point must be within $1/\sqrt{2}$. Or that might be entirely on the wrong track.

Bear with me.
Okay...  So two points $A$ and $D$ must be the furthest (or tied for furthest) apart. Say they are $d$ apart and $d \le 1$.  We might as well assume they are exactly $1$ apart; everything we calculate will be in terms of that distance.
I drew the following image Based on $A$ and $D$:

The Black circles are centered at $A$ and $D$ and have a radius of $1$ (or $d$).  And the red circles have a radius of $\frac 1{\sqrt 2}$.  (As $\frac 12 < \frac 1{\sqrt 2} < 1$ we know the red circles intersect.)
The points $e$ and $f$ are where the circles of radius $\frac 1{\sqrt 2}$ intersect.  And $g$ is are the lines $ef$ and $AD$ intersect.
The geometry you need to know is that $\triangle Aeg$ and $\triangle Afg$ are congruent right triangles and by the Pythagorean theorem that $eg^2 + Ag^2 = Ae^2$.  But $Ae=\frac 1{\sqrt 2}$ and $Ag = \frac 12$ so .... $eg = \frac 12$.
That means the distance from $e$ to $f$ is $1$.
Well... so?
Well where can we put $B$ and $C$?
Well we can't put them outside the black circles.  That'd mean they are more than $1$ away from both $A$ and $D$.
We can't put them in the blue areas because that would mean they are more than one away from either $A$ or from $D$.
We can't put one of them in a light green area the other in the other light green area because that would mean they are more than $1$ apart from each other.
If we put either $B$ or $C$ in a lavender or in the yellow area that would mean that point is within $\frac 1{\sqrt 2}$ of either $A$ or $D$ and our result would be fullfilled.
The only unconsidered option is that both $B$ and $C$ are place in the same light green area.  Then they'd both be more than $\frac 1{\sqrt 2}$ for $A$ and $B$ but they'd be pretty close to each other.
How close? 
