 Solution

The fourth Chebyshev polynomial is $T_4(x)=8x^4-8x^2+1$. Therefore takeIndeed $x^4-p(x)=T_4(x)/8$, and $|T_4(x)|\le 1$ on $[-1,1]$, as follows from $T_4(\cos \theta )=\cos 4\theta$.

Here is a direct Chebyshev alternation theorem argument proving optimality, without assuming the theorem. At the five ordered pointsthe 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.