= Solution
For a real <continuous function> $f$, the <Chebyshev alternation theorem> asserts that a polynomial $p\in\mathcal P_n$ is a <best uniform approximation> exactly when either it is exact or its residual $r=f-p$ has $n+2$ ordered alternating extrema:
$$
\boxed{0\le x_0<\cdots<x_{n+1}\le1,\qquad
r(x_j)=s(-1)^j\|r\|_\infty,\quad s\in\{1,-1\}.}
$$
Here $\mathcal P_n$ means degree at most $n$, as required for a <linear subspace>.
For sufficiency, if a polynomial perturbation $v\in\mathcal P_n$ violated the <Kolmogorov criterion for uniform approximation>, then $r(x_j)v(x_j)>0$ at every displayed point. Thus $v(x_j)$ would alternate signs, forcing a distinct zero in each of the $n+1$ intervals $(x_j,x_{j+1})$ by the <intermediate value theorem>. A nonzero polynomial of degree at most $n$ cannot have that many zeros; the zero polynomial does not violate the criterion either.
For necessity, assume $M=\|r\|_\infty>0$ and no such alternating sequence exists. The two compact sets $E_+=\{r=M\}$ and $E_-=\{r=-M\}$ are disjoint and, when both are nonempty, have positive separation. Hence the extremal set has finitely many successive sign runs: infinitely many alternations would contradict that separation on the bounded interval. If there are $m$ changes of sign between those runs, then selecting one point from each run gives an alternating sequence of length $m+1$, so $m\le n$. Choose a point $z_j$ in the gap between each pair of consecutive opposite-sign runs, and set
$$
v(x)=C\prod_{j=1}^m(x-z_j).
$$
Choose the sign of $C$ so that $v$ agrees with $r$ on the first run. The roots lie outside the extremal set and each changes the sign, so $rv>0$ at every extremal point. Since $\deg v=m\le n$, this contradicts the <Kolmogorov criterion for uniform approximation>. Therefore a best polynomial must have the required alternation.
Existence also holds: a minimizing sequence has bounded <supremum norms>, hence bounded coefficients by evaluating at $n+1$ fixed distinct points and inverting the resulting <Vandermonde matrix>. A convergent subsequence of coefficients yields a minimizer. Finally, if $p$ and $q$ are both best, their average is best; at its alternating extrema both individual residuals must equal the same extremal value. Consequently $p-q$ has at least $n+2$ zeros and is identically zero. \b[The best polynomial exists, is unique, and is characterized by alternation.]
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