= Solution
For real continuous $f$, the <Chebyshev alternation theorem> says that $p\in\mathcal P_n$ is the unique <best uniform approximation> to $f$ precisely when there are $n+2$ ordered points $0\le x_0<\cdots<x_{n+1}\le1$ and a sign $\sigma\in\{-1,1\}$ such that
$$
f(x_j)-p(x_j)=\sigma(-1)^j\|f-p\|_\infty.
$$
Here $\mathcal P_n$ means degree at most $n$. Best approximants exist: a minimizing sequence is bounded in the finite-dimensional <polynomial> space, and <finite-dimensional equivalence of norms> gives a convergent <coefficient> subsequence. When $f=p$, the zero error is treated separately; the derivative-sign hypothesis below excludes that case.
Let $p_n$ be a best degree-at-most-$n$ approximant and suppose for contradiction that $E_n(f)=E_{n+1}(f)$. Since $f^{(n+1)}>0$, $f$ is not a degree-at-most-$n$ <polynomial> and its error is nonzero. The same $p_n$ would be best in $\mathcal P_{n+1}$, so the <Chebyshev alternation theorem> at that degree supplies $n+3$ alternating error extrema. The <intermediate value theorem> gives at least $n+2$ distinct zeros of $e=f-p_n$, one between each consecutive pair of extrema.
Repeated <Rolle's theorem> then forces a zero of $e^{(n+1)}$. But $p_n^{(n+1)}=0$, whereas $e^{(n+1)}=f^{(n+1)}>0$ everywhere. This contradiction proves
$$
\boxed{E_n(f)>E_{n+1}(f).}
$$
The argument proves <strict decrease of best polynomial approximation under a derivative sign>; an everywhere negative $(n+1)$st <derivative> gives the same strict conclusion after replacing $f$ by $-f$.
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