If $X=\Bbb R^n$ is equipped with the usual topology $\tau$ and the usual metric $d$, and we define the function $\pi_U:X\to X$ for every bounded $U\in\tau $ as
$$\pi_U(p)=\int_U d(x,p)dx$$
I need to prove that for every bounded $U\in\tau$ there exists a $p\in X$ such that for all $q\in X$ it holds that $\pi_U(p)<\pi_U(q)$ i.e. that $\pi_U$ has a unique point that attains its minimum value. I need to reach to a contradiction when defining two points with said property, but cannot find any, any help would be appreciated.
An example would be $\pi_X$ where $X$ is the 2d disk or radius $r$(the set of points such that its norm is less than $r$) centered at $(0,0)$, then $(0,0)$ is a point with said property. Same with a rectangle of length $2r$ centered at the origin.
This is not true in the general case. Suppose $n=1$ and $U=(-3,-1)\cup(1,3)$, clearly $U$ is an open bounded subset of $\Bbb R$. Then every value $-1<y<1$ of $\int_Ud(x,y)dx$ will yield the minimum value of the function, $8$, so $\pi_{(-3,-1)\cup(1,3)}$ has not a unique value that attains its minimum value, there are actually uncountable many of them.