= Solution
Let $p\in A$, $r=f-p$, $E=\|r\|_\infty$ and $M=\{x\in[0,1]:|r(x)|=E\}$. The <Kolmogorov criterion for uniform approximation> says
$$
\boxed{p\text{ is best}\quad\Longleftrightarrow\quad\forall h\in A,\ \min_{x\in M}r(x)h(x)\le0.}
$$
If $E=0$, $p=f$ is already best and the criterion holds. Suppose $E>0$. If $q=p+h$ were strictly better, then $|r(x)-h(x)|<|r(x)|=E$ on $M$, which forces $r(x)h(x)>0$ there. Conversely, if a direction $h$ has $rh>0$ everywhere on $M$, <compactness> gives a positive lower bound on this product. On a sufficiently small neighborhood of $M$, $r$ and $h$ therefore have matching signs with $|h|$ bounded away from zero and $|r|$ bounded away from zero. On its compact complement, $|r|\le E-\delta$ for some $\delta>0$. Choose $\varepsilon>0$ so small that $\varepsilon\|h\|_\infty<\delta$ and that subtracting $\varepsilon h$ does not cross zero on that neighborhood. Then $\|r-\varepsilon h\|_\infty<E$, so $p$ is not a <best uniform approximation>. This proves both directions.
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