Let $f$ be a continuous function on $[0,1]$ and for any $n \geq 1$ let us define the polynomial function : $$ \mathcal{P}_{n}(x)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) x^{k}(1-x)^{n-k} f\left(\frac{k}{n}\right) $$
For any $x \in[0,1]$ let $\left(\varepsilon_{n}(x)\right)_{n>1}$ be a sequence of i.i.d. Bernoulii r.v. of parameter $x$. Let then $S_{n}(x)=\varepsilon_{1}(x)+\cdots+\varepsilon_{n}(x)$.
$\mathcal{P}_{n}(x)$ can be interpreted as the following expectation : $$\mathbb{E}[f(\frac{S_n(x)}{n})]$$
By using Bienaymé-Chebychev's inequality we can show that : $$ \forall \varepsilon>0, \quad \mathbb{P}\left(\left|\frac{S_{n}(x)}{n}-x\right| \geq \varepsilon\right) \leq \frac{1}{4 n \varepsilon^{2}} $$
thus $\frac{S_{n}(x)}{n}$ converges in probability and thus in distribution to $x$, consequently $\mathcal{P}_{n}(x)$ converges to $E[f(x)] = f(x)$, I'm not sure how to establish the convergence with the supremum as stated in the title though, any help will be greatly appreciated.
Denote $\zeta_n(x) = S_n(x)/n$. For any $\varepsilon>0$, $$ \bigl|\mathrm{E}[f(\zeta_n(x)] - f(x)\bigr|\le \mathrm{E}\left[\bigl|f(\zeta_n(x))-f(x)\bigr| \right] \\= \mathrm{E}\left[\bigl|f(\zeta_n(x))-f(x)\bigr|\mathbf{1}_{|\zeta_n(x)-x|\le \varepsilon} \right] + \mathrm{E}\left[\bigl|f(\zeta_n(x))-f(x)\bigr|\mathbf{1}_{|\zeta_n(x)-x|> \varepsilon} \right]\\ \le \omega_f(\varepsilon) + 2\sup|f| \cdot \mathrm{P}(|\zeta_n(x)-x|> \varepsilon), $$ where $\omega_f(\varepsilon) = \sup_{0\le s<t<s+\varepsilon\le 1}|f(t) - f(s)|$ is the modulus of coninuity of $f$.
I believe you can conclude from here, using Chebyshev's inequality.