= Solution
Let $u\in\mathcal U$, $r=f-u$, $M=\|r\|_\infty$, and $E=\{x\in[0,1]:|r(x)|=M\}$. The <Kolmogorov criterion for uniform approximation> says
$$
\boxed{u\text{ is a best approximant}
\iff\forall v\in\mathcal U,\quad
\min_{x\in E}r(x)v(x)\le0.}
$$
Equivalently, no available perturbation $v$ has strictly the same sign as the nonzero residual at every extremal point. This statement concerns a candidate $u$ in the <linear subspace>; it does not assume every subspace has an attained best approximation. If $M=0$, $u=f$ is already exact and the criterion is automatic.
To see sufficiency, for any competitor $u+v$, the criterion supplies an $x\in E$ with $r(x)v(x)\le0$. Then $|f(x)-u(x)-v(x)|\ge|r(x)|=M$, so the competitor cannot have smaller <supremum norm> error.
For necessity, suppose $M>0$ and some $v\in\mathcal U$ has $r(x)v(x)>0$ on $E$. Since $E$ is compact, this product has a positive minimum there. By <continuity>, a neighborhood $O$ of $E$ has $rv\ge\eta>0$. For sufficiently small $\varepsilon>0$,
$$
|r-\varepsilon v|^2=r^2-2\varepsilon rv+\varepsilon^2v^2<M^2\quad\text{on }O.
$$
On the compact complement of $O$, $|r|$ has a strict gap below $M$, and taking $\varepsilon\|v\|_\infty$ smaller than this gap preserves it. Hence $u+\varepsilon v$ improves the uniform error everywhere, contradicting <best uniform approximation>.
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