Proving Integrability Using Uniform Continuity 
Prove that if a function $f(x)$ is Uniformly Continuous on $[a, b]$ then it is Integrable on $[a, b]$.

As the function $f(x)$ is Uniformly continuous on $[a, b],$ so for a given $\epsilon$ there exists a $\delta_0$ such that for $|x-y|\leq \delta_0$ we have $$|f(x)-f(y)|<\frac{\epsilon}{b-a}.$$ Now choose a partition of the interval $[a, b]$ whose each sub interval has the width $\delta_0$ and hence number of sub intervals equals $n=\frac{b-a}{\delta_0}$ and hence we have $$U-L={\delta_0}\sum_{k=1}^{n}(f(M_k)-f(m_k)<\frac{b-a}{\delta_0}*\delta_0*\frac{\epsilon}{b-a}=\epsilon,$$ where $M_k, m_k$ denote the maximum and minimum values of $f(x)$ on the $k^{th}$ sub interval respectively.
Is my proof correct?
 A: $\frac {b-a} {\delta_0}$ need not be an integer. Take any positive integer $n > \frac {b-a} {\delta_0}$ and divide the interval into $n$ equal parts. Then the length of each subinterval is less than $\delta_0$ so your inequalities are true. So $U-L <\epsilon$ for this and any finer partition which gives Riemann integrability. 
A: yes, your proof is correct.But you have to cleary justify that $\sup$ and $\inf$ becomes $\max$ and $\min$ because of being continuous on compact set.
Or 
$$\sup_{x \in S} |f(x) - f (y)| = \sup_{x \in S} f - \inf_{x \in S} f $$
use above fact and conclude what we want. since we know $|f(x) - f(y)| < \epsilon \implies \epsilon  \text{ is upper bound} \implies \sup_{x \in S} f - \inf_{x \in S} f \leq \epsilon$ . But then you have to adjust for required $\epsilon$ because it is not strict inequality anymore.
Also just choose partition s.t $(x_{i+1} - x_{i} )< \delta , \forall i$ you don't need to divide into uniform intervals.
$$U - L = \sum_{i=1}^n \Big[ \sup_{x \in [x_i , x_{i+1}]} f(x) - \inf_{x \in[x_i,x{i+1}]} f(x) \Big] (x_{i+1} - x_{i}) < \frac{\epsilon}{b-a} \sum_{i=1}^n (x_{i+1} - x_{i}) = \epsilon$$
