Prove the Supremum is attained. Let $F$ denote denote the set of real valued functions on $[0,1]$ such that,
1) $ \; |f(x)| \leq 1 \; \forall x \; \in [0,1]$
2) $ \; |f(x)-f(x')| \leq |x-x'| \; \: \forall x,x' \: \in [0,1] $
Prove that that the following supremum is attained. $$\sup_{f \in F} \int_0^1 f(x) \sin(\frac{1}{x})\,dx$$ 
Thoughts :
Conditions 1) and 2) jointly imply that $F \subset C[0,1]$ is Equicontinuous and bounded. Hence by the Arzela Theorem,  $F$ has compact closure. In this case I think $F$ is closed although I'm not sure how to show this.
For each $n  \in \mathbb{N} \; \exists \; \; f_n(x) \in F$ such that, $$\sup_{f \in F} \int_0^1 f(x) \sin(\frac{1}{x})\,dx -\frac{1}{n} \: < \int_0^1 f_n(x) \sin(\frac{1}{x})\,dx \leq \; \:\sup_{f \in F} \int_0^1 f(x) \sin(\frac{1}{x})\,dx$$
Then $f_n(x)$ is a sequence in $F$ and so has a convergent subsequence $f_{n_{k}}(x) \rightarrow f$ by compactness.
I think I've almost got the answer except I can't justify that $$\lim_{k\rightarrow \infty} \int_0^1 f_{n_{k}}(x) \sin(\frac{1}{x})\,dx = \int_0^1 f(x) \sin(\frac{1}{x})\,dx$$
I've seen this work in Lebesgue theory ( Dominated Convergence ) but not sure how it really works with Riemann integrals.
 A: As you already noted, $F$ is equicontinuous and (pointwise) bounded, so Arzela-Ascoli implies its closure is compact. Let's verify $F$ is closed: Suppose $f_n$ is a sequence in $F$ which converges uniformly to some continuous function $g$. Then $f_n$ converges pointwise, hence
\begin{align*}
1)&|g(x)|=\lim |f_n(x)|\leq 1;\qquad\text{and}\\
2)&|g(x)-g(x')|=\lim|f_n(x)-f_n(x')|\leq|x-x'|;
\end{align*}
so $g\in F$. Therefore $F$ is compact (with respect to the uniform norm $\Vert f\Vert_\infty=\sup_x|f(x)|$).
Let's verify that the function $f\in F\mapsto \int_0^1 f(x)\sin(1/x)dx$ is continuous (with respect to the uniform norm). Indeed, for all $f$ and $g$ in $F$,
\begin{align*}
|\int_0^1 f(x)\sin(1/x)d-\int_0^1 g(x)\sin(1/x)dx|&\leq\int_0^1|f(x)-g(x)||\sin(1/x)|dx\\
&\leq\int_0^1|f(x)-g(x)|dx\\
&\leq\int_0^1\Vert f-g\Vert_\infty dx=\Vert f-g\Vert_\infty,
\end{align*}
so in fact the map $f\mapsto \int_0^1f(x)\sin(1/x)dx$ is Lipschitz, hence continuous.
Since a continuous function attains its supremum in a compact, we are done.
