I am having some trouble understanding how the part underlined is obtained.
For convenience let's denote $\frac{1}{n}\Sigma_{i=1}^nU_i\circ1_{[0,p]}$ as x. So The solution tries to apply uniform continuity on this f.
I could not figure out where the term after epsilon comes from in $|f(x)-f(p)|\leq \epsilon + 2||f||*1_{|x-p|\geq \delta}$.
Any help is appreciated!
Notice that
$$|f_n(p)-f(p)|=\left | \sum_{k=0}^n (f(k/n)-f(p)) {n \choose k} p^k (1-p)^{n-k} \right | \\ \leq \sum_{k=0}^n |f(k/n)-f(p)| {n \choose k} p^k (1-p)^{n-k}.$$
Now you split the sum between $\{ k : |k/n-p|<\delta \}$ and $\{ k: |k/n-p| \geq \delta \}$. In the first part, $|f(k/n)-f(p)|<\varepsilon$ and the weights of those $k$'s sum to at most $1$, so that term is at most $\varepsilon$.
In the second part, you have no control over $|f(k/n)-f(p)|$, so you need a bound on $\sum_{k : |k/n-p| \geq \delta} {n \choose k} p^k (1-p)^{n-k}$ instead. This can be obtained from Chebyshev's inequality: the standard deviation of a binomial($n,p$) variable divided by $n$ is $\sqrt{p(1-p)/n}$ so Chebyshev's inequality gives a bound for this sum of $\frac{p(1-p)}{n\delta^2}$. Now it gets multiplied by $2 \| f \|$, so
$$|f_n(p)-f(p)| \leq \varepsilon + \frac{2 \| f \| p(1-p)}{n\delta^2}.$$