Proof that f is Riemann Integrable 
Theorem 6.1.8a: If $f$ is continuous on $[a,b]$ then $f$ is Riemann Integrable on $[a,b]$.
Theorem 6.1.7:  A bounded real-valued function $f$ is Riemann Integrable on $[a,b]$ if and only if for every $e > 0$, there exists a partition $P$ of $[a,b]$ such that $U(\mathcal{P},f) - L(\mathcal{P},f) < e$. Futhermore, if $P$ is a partition of $[a,b]$ for which the above inequality holds, then the inequality also holds for all refinements on $P$. 
Theorem 6.1.13: A bounded real-valued function f of $[a,b]$ is a Riemann Integrable if and only if the set of discontinuities of f has zero measure.
So we get any abstract partition of $[a,c-d]$ and one of $[c+d,b]$ such that $U[\mathcal{P}_1,f]-L[\mathcal{P}_2,f] < \frac{e}{3}$?
Now I get confused as to how to define $d$. 
Please correct/help! 

Since $f$ is continuous on $[a, c - \delta]$ and $[c + \delta, b]$ therefore it is integrable on both these intervals.
Let $P_{1}$ be partition of $[a, c - \delta]$ and $P_{2}$ be partition of $[c + \delta, b]$ such that $U(P_{1}, f) - L(P_{1}, f) < \epsilon / 3$ and $U(P_{2}, f) - L(P_{2}, f) < \epsilon / 3$. Now choose $P_{3} = \{c - \delta, c + \delta\}$ as a partition of $[c - \delta, c + \delta]$.
The trick here is to choose $\delta$ small enough such that $U(P_{3}, f) - L(P_{3}, f) < \epsilon / 3$. Clearly we have $$U(P_{3}, f) - L(P_{3}, f) = \{c + \delta - (c - \delta)\}(M_{c} - m_{c}) = 2\delta(M_{c} - m_{c})$$ where $M_{c} = \sup \{f(x)\mid x \in [c - \delta, c + \delta]\}$ and $m_{c} = \inf \{f(x)\mid x \in [c - \delta, c + \delta]\}$. Clearly $M_{c} - m_{c} \leq 2M$ where $M = \sup \{|f(x)|\mid x \in [a, b]\}$ and therefore if we choose $\delta < \epsilon / 12M$ then $$U(P_{3}, f) - L(P_{3}, f) = 2\delta(M_{c} - m_{c}) < 2\cdot\frac{\epsilon}{12M}\cdot 2M = \frac{\epsilon}{3}$$
Let $P = P_{1} \cup P_{2} \cup P_{3}$ then $P$ is a partition of $[a, b]$ such that
$\displaystyle \begin{aligned}U(P, f) - L(P, f) &= U(P_{1}, f) - L(P_{1}, f) + U(P_{2}, f) - L(P_{2}, f) + U(P_{3}, f) - L(P_{3}, f)\\
&< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}\\
&= \epsilon\end{aligned}$
Hence $f$ is Riemann-Integrable on $[a, b]$.
