 Solution

Let $p\in A$, $r=f-p$, $E=\Vert r{\Vert }_{\infty }$ and $M=\{x\in [0,1]:|r(x)|=E\}$. The Kolmogorov criterion for uniform approximation saysIf $E=0$, $p=f$ 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 $\epsilon >0$ so small that $\epsilon \Vert h{\Vert }_{\infty }<\delta$ and that subtracting $\epsilon h$ does not cross zero on that neighborhood. Then $\Vert r-\epsilon h{\Vert }_{\infty }<E$, so $p$ is not a best uniform approximation. This proves both directions.