= Solution
The <Chebyshev alternation theorem> says that a <polynomial> $P$ of <degree of a polynomial>[degree] at most $n$ is a <best uniform approximation> to $f\in C([0,1])$ if and only if there are $n+2$ points
$$
0\leq t_0<t_1<\cdots<t_{n+1}\leq1
$$
and a sign $\varepsilon\in\{1,-1\}$ for which
$$
f(t_j)-P(t_j)=\varepsilon(-1)^j\lVert f-P\rVert_\infty.
$$
This is the equiripple criterion.
To prove sufficiency, put $E=\lVert f-P\rVert_\infty$ and suppose that another polynomial $Q$ of degree at most $n$ satisfies $\lVert f-Q\rVert_\infty<E$. At every $t_j$, the difference $R=Q-P$ must have the same sign as $f-P$; hence its signs alternate at the $n+2$ ordered points. The <intermediate value theorem> then gives at least one distinct root of $R$ in every interval $(t_j,t_{j+1})$, for at least $n+1$ roots in all. The <Lagrange root bound over a field> forces the nonzero polynomial $R$ to have degree at least $n+1$, contradicting $\deg R\leq n$. Therefore no $Q$ gives a smaller error, and $P$ is best.
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