One solution is:
Note that equality holds for a multiple of the n-th Chebyshev polynomial $T_{n}(X)$ The leading coefficient of $T_{n}$ equals $2^{n-1}$, so $C_{n}(X) = \frac{1}{2^{n-1}}T_{n}(X)$ is a monic polynomial and
$$ |T_{n}(X)| = \frac{1}{2^{n-1}}|\cos{(n\arccos{x}})| \leq \frac{1}{2^{n-1}} \ \ \ \ za \ \ x \in [-1,1] $$
Moreover the values of $T_{n}$ at points $1, \cos{\frac{\pi}{n}}, \cos{\frac{2\pi}{n}}, \dots , \cos{\frac{(n-1)\pi}{n}}, -1$ are alternately $\frac{1}{2^{n-1}}$ and $-\frac{1}{2^{n-1}}$
Now suppose that $P \neq T_{n}$ is a monic polynomial such that $\max_{-1\leq x \leq 1} |P(x)| < \frac{1}{2^{n-1}}$ Then $P(X)-C_{n}(X)$ at points $1, \cos{\frac{\pi}{n}}, \dots , \cos{\frac{(n-1)\pi}{n}}, -1$ alternately takes positive and negative values. Therefore the polynomial $P-C_{n}$ has at least $n$ zeros, namely, at least one is very interval between two adjacent points. However, $P-C_{n}$ is a polynomial of degree $n-1$ as the monimial $x^n$ is canceled, so we have arrived at a contradition.
Now, I'm looking for other solutions to prove it, please comment on