= Solution
Fix the degree bound $n$. Existence can be proved without a general uniqueness theorem: a minimizing sequence in $\mathcal P_n$ is uniformly bounded because its distances from $f$ are bounded. Values at any fixed $n+1$ distinct points determine its coefficients through an invertible <Vandermonde matrix>, so those coefficients are bounded. A convergent subsequence supplies a <polynomial> attaining the minimum.
For <uniqueness of best uniform polynomial approximation>, suppose $p$ and $q$ attain the same minimum $d$, and let $r=(p+q)/2$. The <triangle inequality> makes $r$ another minimizer. If $d=0$, both <polynomials> equal $f$, so assume $d>0$. Put $e=f-r$ and $E=\{|e|=d\}$. At each $x\in E$, the two real errors $f-p$ and $f-q$ lie in $[-d,d]$ and their average equals an endpoint. Hence they are equal there, and $p(x)=q(x)$.
The set $E$ must contain at least $n+2$ points. Otherwise interpolate the values $\operatorname{sign}e(x)$ on its at most $n+1$ points by a <polynomial> $v\in\mathcal P_n$. Then $ev>0$ on $E$. Continuity gives this positivity on a neighbourhood of $E$, while the error has a strict gap below $d$ on the compact complement. A sufficiently small positive multiple of $v$ decreases the maximum error, contradicting optimality of $r$. This is an elementary perturbation argument, not an appeal to Haar's theorem.
Thus $p-q$ has at least $n+2$ distinct zeros. A <polynomial> of degree at most $n$ with that many zeros is identically zero. \b[The best uniform <polynomial> is unique for every fixed degree bound]: $\boxed{p=q}$.
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