Continuity implies Riemann integrability 
Let $X$ be a Banach space and $f:[a,b] \to X$ be continuous. Show that $f$ is Riemann integrable.

My attempt:
Let $t_i^n=a+i\frac{b-a}{n},(i=1,2,\cdots,n)$ and $$s_n=\sum_{i=1}^n f(t_i^n) (t_i^n-t_{i-1}^n)=\sum_{i=1}^n f(t_i^n)\frac{b-a}{n}$$
It suffices to show that $(s_n)$ is Cauchy. For $n,m \in \mathbb N$ with $n>m,$
\begin{align*}
s_n-s_m &= \sum_{i=1}^n f(t_i^n)\frac{b-a}{n}-\sum_{i=1}^m f(t_i^m)\frac{b-a}{m}\\
&<(b-a)\left[\frac{1}{m} \sum_{i=1}^m \left(f(t_i^n)-f(t_i^m)\right)+\frac1n \sum_{i=m+1}^n f(t_i^n)\right]
\end{align*}
Since, $f$ is continuous on a compact set, therefore it is uniformly continuous and bounded. Therefore for given $\epsilon >0,$ there exists $\delta>0$ such that $$|x-y|<\delta \implies \|f(x)-f(y)\|<\epsilon.$$
For $i\in \{1,2,\cdots,m\},$ consider $$t_i^n-t_i^m=i(b-a)\left(\frac 1n-\frac 1m\right)\to 0 \text{ as }n,m \to \infty$$Thus, there exists $n_0\in \mathbb N$ such that $|t_i^n-t_i^m|<\delta$ for all $n,m \geq n_0.$ It follows that
$$\|s_n-s_m\|<(b-a)\left(\epsilon+\frac{n-m}{n}\|f\|_\infty\right)$$ for all $n,m \geq n_0.$
From here, if I can show that $\frac{n-m}{n} \to 0$ as $n,m \to \infty ,$ then I'll be done. 
Edit: Since,$$\lim_{m \to \infty } \lim_{n \to \infty}\frac{n-m}{n}=1\neq\lim_{n \to \infty } \lim_{m \to \infty}\frac{n-m}{n}$$
So, $\lim_{m,n \to \infty}\frac{n-m}{n}$ doesn't exist. So my proof fails.

How should I approximate the second term in the summation to prove the sequence is Cauchy? 

 A: Since $[a,b]$ is compact, $f$ is uniformly continuous and for any $\epsilon > 0$ there exists $\delta >0$ such that if $|x - y| < \delta$, then $\|f(x) - f(y)\| < \frac{\epsilon}{2(b-a)}$.
Define a partition $P_n = (t_0^n, t_1^n, \ldots, t_n^n)$ where  $t_i^n = a + i(b-a)/n$ and a sequence of Riemann sums $S_n = \sum_{i=1}^n f(t_i^n)(t_i^n - t_{i-1}^n)$. Take $n$ such that $(b-a)/n < \delta$.
Suppose $R$ is any refinement of $P_n$. Any subinterval $[t_{i-1}^n, t_{i}^n]$ used in  $S_n$ can be decomposed as a union of subintervals of $R$,
$$[t_{i-1}^n,t_{i}^n] = \bigcup_{k=1}^{n_i}[y_{i,k-1}, y_{i,k}] .$$
Now take any Riemann sum $S(R,f)$ for the refined partition with tags $\eta_{i,k} \in [y_{i,k-1}, y_{i,k}] \subset[t_{i-1}^n,t_{i}^n].$ Clearly $|\eta_{i,k} - t_i^n| < \delta$ and 
$$\begin{align}\left\|S_n - S(f,R)\right\| &= \left\|\sum_{i=1}^n f(t_i^n)(t_i^n - t_{i-1}^n) - \sum_{i=1}^n \sum_{k=1}^{n_i}f(\eta_{i,k})(y_{i,k} - y_{i,k-1})\right\| \\ &\leqslant \sum_{i=1}^n \sum_{k=1}^{n_i}\|f(t_i^n) - f(\eta_{i,k})\|(y_{i,k} - y_{i,k-1}) \\ &\leqslant \sum_{i=1}^n \sum_{k=1}^{n_i}\frac{\epsilon}{2(b-a)}(y_{i,k} - y_{i,k-1}) \\ &= \epsilon/2\end{align}$$
Now consider the sum $S_m$ corresponding to the partition $P_m$ where $m > n$.
Let $R = P_n \cup P_m$ be a common refinement of both $P_n$ and $P_m$.  
Since $\|P_m\| < \|P_n\| < \delta$  we have $\|S_m - S(f,R)\| < \epsilon/2$ by the preceeding argument, and it follows that
$$\|S_m - S_n\| \leqslant \|S_m - S(f,R)\| + \|S_n - S(f,R)\| < \epsilon/2 + \epsilon/2 = \epsilon.$$ 
Therefore, $(S_n)$ is a Cauchy sequence.
