The problem
Define the Bernstein polynomial of $f$ by
$$B_nf(x) = \sum_{i=0}^n\binom n i x^i(1-x)^{n-i} f(i/n)$$
Prove without the use of the Stone-Weierstrass theorem that $\limsup_{n\to\infty}|B_nf - f| = 0$ where $f$ is any continuous function on $[0,1]$.
Hint: For all $\varepsilon>0$ there is a constant $\alpha$, dependent only only $f$ and $\varepsilon$, such that for all $x,y\in[0,1]$ we have
$$ |f(x)-f(y)| < \varepsilon + \alpha M(x-y)^2 $$
where $M=\sup_{x\in[0,1]}|f(x)|$.
My attempt
We have that $f$ is uniformly continuous on $[0,1]$, and so is $B_nf$. So if we let $\varepsilon > 0$ there is some $\delta>0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. We could further have that $|B_nf(x)-B_nf(y)|<\varepsilon$ whenever $|x-y|<\delta$ if we wanted to -- I'm not sure of the value of this, but it can't hurt to have it.
Also,
$$ \sup_{x\in[0,1]} |B_nf(x) - f(x)| \le \sup_{x\in[0,1]} (|B_n f(x)|+|f(x)|) \le \sup_{x\in[0,1]}(B_n|f(x)|+|f(x)|) $$
I don't know if that's of any value either.
I am especially struggling to see the relevance of the hint. I've found the value of $\alpha$ and I doubt its exact definition is relevant to this problem.