= Solution
The fourth <Chebyshev polynomial> is $T_4(x)=8x^4-8x^2+1$. Therefore take
$$
\boxed{p(x)=x^2-\frac18,\qquad \|x^4-p(x)\|_\infty=\frac18.}
$$
Indeed $x^4-p(x)=T_4(x)/8$, and $|T_4(x)|\le1$ on $[-1,1]$, as follows from $T_4(\cos\theta)=\cos4\theta$.
Here is a direct <Chebyshev alternation theorem> argument proving optimality, without assuming the theorem. At the five ordered points
$$
-1,\quad-1/\sqrt2,\quad0,\quad1/\sqrt2,\quad1,
$$
the error $x^4-p(x)$ takes the alternating values $1/8,-1/8,1/8,-1/8,1/8$. If a <polynomial> $q$ of degree at most three had strictly smaller uniform error, then $q-p$ would have alternating strictly positive and strictly negative values at these points. The <intermediate value theorem> would give a distinct root in each of the four intervening intervals. A nonzero <polynomial> of degree at most three cannot have four distinct roots. This contradiction proves the required minimax inequality.
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