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We work in the usual real-valued setting of the Chebyshev alternation theorem. For $f\in C[-1,1]$ and an algebraic polynomial $p$ of degree at most $n$, the theorem says that $p$ is a best uniform approximation if and only if its error has $n+2$ ordered points with alternating maximal values:If $f=p$, the zero-error case is included directly.

Existence follows, for example, by taking a minimizing sequence: its supremum norms are bounded, all norms on the finite-dimensional polynomial space are equivalent, and a convergent coefficient subsequence attains the infimum. To prove uniqueness of best uniform polynomial approximation, let $p,q$ both attain the minimum error $E$. Their average $r=(p+q)/2$ has error at most $E$ by the triangle inequality and therefore exactly $E$ by minimality.

If $E=0$, both polynomials equal $f$ and are equal. Otherwise apply the Chebyshev alternation theorem to $r$. At every alternating extremal point, $f-r$ is either $E$ or $-E$. But it is the average of $f-p$ and $f-q$, each lying in $[-E,E]$. An average attains an endpoint of this interval only when both entries equal that endpoint. Hence $p(x_j)=q(x_j)$ at all $n+2$ points. The difference is a degree-at-most-$n$ polynomial with more than $n$ distinct zeros, so it is identically zero. The best approximating polynomial is unique.

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 $\mathrm{sign}e(x)$ 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. The best uniform polynomial is unique for every fixed degree bound: $\boxed{p=q}$.