Prove there exists a point p in P so that there are $u\neq v\in S$ with $d(p,u)=d(p,v)=\max_{x\in S} d(p,x),$ 
Let $S$ be a finite set of at least two points in the xy-plane, denoted by P. Prove there exists a point p in P so that there are $u\neq v\in S$ with $d(p,u)=d(p,v)=\max_{x\in S} d(p,x),$ where $d(x,y)$ is the Euclidean distance between x and y.

I'm not sure if the extremal principle is useful here. Assume for a contradiction that for every point p in P, its maximum distance to S is unique (obviously the maximum must be achieved for at least one point in S). I know the triangle inequality for distance, which says $d(a,b) \leq d(a,c)+d(c,b)$ for all $a,b\in \mathbb{R}^2$. There might be a way to constructively choose a point p depending on S.
 A: Consider a circle that encloses all points in $S$. Slowly shrink it until it touches one point of $S$, then carefully tighten it further so that it touches another point but all remaining points are still inside. Call them $u$ and $v$. (This procedure can probably be formalized using the convex hull of $S$, or something.) Now draw a perpendicular bisector between the two points on the circle, and choose any point $p$ on this line outside the circle ($p$ should be closer to the diameter than to the chord $uv$). Verify that $p$ satisfies the properties you ask for.
EDIT: That was a very obtuse answer, I think I have an improvement! The first part is the same, find a circle that encloses all points but with exactly two points ($u$ and $v$) on the circumference. Then you can just choose $p$ as the centre of this circle, and you're done!
As for why such a circle exists, we can agree that we can draw a circle touching only one point and being so big that it encloses all other points (e.g. in the limit of being a straight line). If we shrink it further, eventually it will start enclosing some points. To be able to do that, it had to touch some other point first. Simply stop the shrinking process when that touching happens. I'm sorry I don't know how to make this rigorous though :'(
A: Let's assume by contradiction that for exery $p \in \mathbb R^2$ there exsit a unique point $s_p \in S$ such that $d(p,s_p)=\max_{x\in S} d(p,x)$.
define a function $f:\mathbb R^2 \to S$ by $f(p) = s_p$
now pick an arbitrary point p in $\mathbb R^2$
define $\gamma:[0,1] \to \mathbb R^2$ by $\gamma(t)=ts_p+(1-t)p$. Notice that $\gamma$ is continous.
and look at $t_0 = \sup\{t \in [0,1] | f(\gamma(t))=s_p\}$
[notice that $0 \in \{t \in [0,1] | f(\gamma(t))=s_p\}$ so it is well defined]
now we will prove that $f(t_0) \neq s_p$. let's assume otherwise  $f(t_0)=s_p$

*

*(if $t_0<1$): for every $s_p \neq s \in S$ we have
$d(s_p,\gamma(t_0))>d(s,\gamma(t_0))$ now from continouty of these functions we can find $\delta_s>0$ such that $[t_0,t_0+\delta_s) \in [0,1]$ and for each $t \in [t_0,t_0+\delta_s]$ we have that $d(s_p,\gamma(t))>d(s,\gamma(t))$. Now we can define $\delta = \min_{s_p \neq s\in S} \delta_s$, we have that $\delta$ exsit and greater than 0 as S is finite. and thus for $t=t_0+\delta$ we have that $d(s_p,\gamma(t))=\max_{x\in S} d(\gamma(t),x)=f(\gamma(t))$
and thus $t\in \{t \in [0,1] | f(\gamma(t))=s_p\}$ but $t>t_0$ which is contradiction.


*(If $t_0$=1): notice that $\gamma(1)=s_p$ and thus $f(\gamma(1))=f(s_p) \neq s_p$ because $S$ has at least 2 points and the thus has points with positive distance from s_p. thus we have $d(f(\gamma(1)),\gamma(1))>d(s_p,\gamma(1))$ and again from continouty we have $\delta>0$ such that for each $t\in [1,1-\delta)$ $d(f(\gamma(1)),\gamma(t))>d(s_p,\gamma(t))$ and thus from the defenition of $f$ we get that$ f(\gamma(t)) \neq s_p$, but that is a contradiction as we get that $1 = \sup\{t \in [0,1] | f(\gamma(t))=s_p\} \le 1-\delta$
and thus we have that $f(t_0) \neq s_p$ and thus $d(f(\gamma(t_0)),\gamma(t_0))>d(s_p,\gamma(t_0))$ and again from continouty we have $\delta>0$ such that for each $t\in [t_0,t_0-\delta)$ $d(f(\gamma(t_0)),\gamma(t))>d(s_p,\gamma(t))$ and thus from the defenition of $f$ we get that$ f(\gamma(t)) \neq s_p$, but that is a contradiction as we get that $t_0 = \sup\{t \in [0,1] | f(\gamma(t))=s_p\} \le t_0-\delta$
and thus we obtain a contradiction.
