Let $S = \{ p\in C([0,1]), p\ \text{polynomial of degree} \leq d: \max |p(x)| \leq 1 \}$
I want to show that $S$ is equicontinuous. Using the definition of equicontinuity we need for all $\epsilon >0$ exists $\delta>0$ for all $x,y \in [0,1],$ for all $p\in S $: for $d(x,y)<\delta$ we have $ |p(x) - p(y)| < \epsilon$
However I cannot bound the difference: $|\sum_{i=0}^{n} c_ix^i - \sum_{i=0}^{n} c_iy^i|=|\sum_{i=0}^{n} c_i(x^i-y^i)|$. We have that $|x-y|<\delta$ so $|x^i-y^i|$ is also bounded, but I think I also need a bound for $\max_{i} |c_i|$ which I cannot find. I think that since $p(x)$ is bounded, $\max_{x\in[0,1]} p(x)$ we should be able to get an upper bound for $\max_i |c_i|$.
I would like to prove this using only the definition of equicontinuity.
Yes, $S$ is equicontinuous. Let $q(x)=p((x+1)/2)$ and consider the Markov brothers' inequality for $k=1$: $$\sup_{x\in [0, 1]}|p'(x)|=2\sup_{x\in [-1, 1]}|q'(x)| \le 2d^2 \sup_{x\in [-1, 1]}|q(x)| \leq 2d^2.$$
P.S. The set $S$ is also closed: if a sequence $p_n(x)=\sum_{k=0}^da_k(n)x^k$ converges uniformly to $f$ in $[0,1]$ then $\max_{x\in [0,1]} |f(x)| \leq 1$ and $(a_k(n))_n$ is a convergent sequence for $k=0,\dots, d$ (use the fact that $(p_n(k/d))_n$ is convergent for $k=0,\dots,d$ and the Vandermonde determinant).