From comments to my question I have formulated my own proofs. If anyone can confirm their correctness (especially on the second one!) I would greatly appreciate it!
Prove that if $f$ is bounded on $[a, b]$ and has exactly one discontinuity in $[a, b]$ then $f$ is Riemann-integrable on $[a, b]$.
Suppose $f$ is bounded on $[a, b]$. Then by definition there exists $M > 0$ such that $\vert f(x) \vert \leq M$ for all $x \in [a, b]$. Suppose $f$ has exactly one discontinuity in $[a, b]$ call it $c$. Suppose $c \in (a, b)$. Let $\varepsilon > 0$. Choose $\delta > 0$ such that $\displaystyle \delta = \frac{\varepsilon}{12M}$ and also we must restrict delta such that $a < c - \delta$ and $c + \delta < b$. Observe that $f$ is continuous on $[a, c - \delta]$ and $[c + \delta, b]$. By Theorem 6.2 $f$ is Riemann-integrable on $[a, c - \delta]$ and $[c + \delta, b]$. By Theorem 6.1 there exists a partition $P_1$ of $[a, c - \delta]$ with $U(P_1, f) - L(P_1, f) < \varepsilon/3$. Again, by Theorem 6.1 there exists a partition $P_2$ of $[c + \delta, b]$ with $U(P_2, f) - L(P_2, f) < \varepsilon/3$. Define $P = P_1 \cup P_2$. Now observe that
\begin{align*}
U(P, f) &= U(P_1, f) + 2\delta \cdot \sup_{x \in [c - \delta, c+ \delta]} f(x) + U(P_2, f) \\
&\leq U(P_1, f) + 2\delta \cdot M + U(P_2, f)
\end{align*}
and
\begin{align*}
L(P, f) &= L(P_1, f) + 2\delta \cdot \inf_{x \in [c - \delta, c+ \delta]} f(x) + L(P_2, f)\\
& \geq L(P_1, f) + 2\delta \cdot (-M) + L(P_2, f)
\end{align*}
which is implies $$-L(P, f) \leq - L(P_1, f) + 2\delta \cdot M - L(P_2, f) $$ Hence we have
\begin{align*}
U(P, f) - L(P, f) &\leq U(P_1, f) + 2\delta \cdot M + U(P_2, f) - L(P_1, f) + 2\delta \cdot M - L(P_2, f)\\
&= \big[U(P_1, f) - L(P_1, f) \big] + 4 \delta \cdot M + \big[U(P_2, f) - L(P_2, f)\big]\\
&< \frac\varepsilon3 + 4M\frac{\varepsilon}{12M} + \frac\varepsilon3\\
&= \frac\varepsilon3 +\frac\varepsilon3 +\frac\varepsilon3 \\
&= \varepsilon
\end{align*}
Thus by Theorem 6.1 we can conclude that $f$ is Riemann-integrable on $[a, b]$.
Now consider $c = b$. Since $f$ is bounded, there exists $M > 0$ such that $\vert f(x) \vert \leq M$ for all $x \in [a, b]$. Let $\varepsilon > 0$. Choose $\delta > 0$ such that $\delta = \frac{\varepsilon}{4M}$ and we must restrict delta such that $a < b - \delta$. Observe that $[a, b - \delta]$ is continuous. By Theorem 6.2 $f$ is Riemann-integrable on $[a, b - \delta]$. Since $f$ is Riemann-integrable on $[a, b - \delta]$, by Theorem 6.1 there exists a partition $P_1$ of $[a, b - \delta]$ such that $U(P_1, f) - L(P_1, f) < \frac\varepsilon2$. Now consider the partition $P = P_1 \cup \{b\}$ and observe that $\displaystyle U(P, f) = U(P_1, f) + \delta \sup_{x \in [b - \delta, b]} f(x) \leq U(P_1, f) + \delta M$. Also $\displaystyle L(P, f) = L(P_1, f) + \delta \inf_{x \in [b - \delta, b]} f(x) \geq L(P_1, f) + \delta(- M)$ which is equivalent to $-L(P, f) \leq -L(P_1, f) + \delta M$. Thus we have
\begin{align*}
U(P, f) - L(P, f)&\leq U(P_1, f) + \delta M -L(P_1, f) + \delta M\\
&= \big[U(P_1, f) - L(P_1, f) \big] + 2\delta M\\
&< \frac\varepsilon2 + 2M \frac{\varepsilon}{4M}\\
&= \frac\varepsilon2 + \frac\varepsilon2\\
&= \varepsilon
\end{align*}
Thus by Theorem 6.1 we can conclude that $f$ is Riemann-integrable on $[a, b]$.
Now consider $c = a$. Since $f$ is bounded, there exists $M > 0$ such that $\vert f(x) \vert \leq M$ for all $x \in [a, b]$. Let $\varepsilon > 0$. Choose $\delta > 0$ such that $\delta = \frac{\varepsilon}{4M}$ and we must restrict delta such that $a + \delta < b$. Observe that $[a + \delta, b]$ is continuous. By Theorem 6.2 $f$ is Riemann-integrable on $[a + \delta, b]$. Since $f$ is Riemann-integrable on $[a + \delta, b]$, by Theorem 6.1 there exists a partition $P_1$ of $[a + \delta, b]$ such that $U(P_1, f) - L(P_1, f) < \frac\varepsilon2$. Now consider the partition $P = P_1 \cup \{a\}$ and observe that $\displaystyle U(P, f) = U(P_1, f) + \delta \sup_{x \in [a, a + \delta]} f(x) \leq U(P_1, f) + \delta M$. Also $\displaystyle L(P, f) = L(P_1, f) + \delta \inf_{x \in [a, a + \delta]} f(x) \geq L(P_1, f) + \delta(- M)$ which is equivalent to $-L(P, f) \leq -L(P_1, f) + \delta M$. Thus we have
\begin{align*}
U(P, f) - L(P, f)&\leq U(P_1, f) + \delta M -L(P_1, f) + \delta M\\
&= \big[U(P_1, f) - L(P_1, f) \big] + 2\delta M\\
&< \frac\varepsilon2 + 2M \frac{\varepsilon}{4M}\\
&= \frac\varepsilon2 + \frac\varepsilon2\\
&= \varepsilon
\end{align*}
Thus by Theorem 6.1 we can conclude that $f$ is Riemann-integrable on $[a, b]$.
Prove that if $f$ is bounded on $[a, b]$ and $f$ has only finitely many discontinuities in $[a, b]$ then $f$ is Riemann-integrable on $[a, b]$.
We will prove this using the principle of mathematical induction. Observe that the case when there is one discontinuity has already been shown. Suppose that $f$ is bounded on $[a, b]$ and $f$ has $n$ discontinuities in $[a, b]$ and suppose $f$ is Riemann-integrable on $[a, b]$. Consider the case when $f$ is bounded on $[a, b]$ and $f$ has $n + 1$ discontinuities in $[a, b]$. Choose the right most discontinuity and call it $c$. Choose $\delta > 0$ such that $[a, c - \delta]$ has $n$ discontinuities. By the inductive hypothesis we can conclude that $f$ is Riemann-integrable on $[a, c - \delta]$. Let $\varepsilon > 0$. Since $f$ is Riemann-integrable on $[a, c - \delta]$, by Theorem 6.1, there exists a partition $P_1$ of $[a, c - \delta]$ with $U(P_1, f) - L(P_1, f) < \varepsilon/2$. Now observe that $f$ is bounded on $[c - \delta, b]$ and by construction has only one discontinuity on $[c - \delta, b]$. By Exercise 6.5 we can conclude that $f$ is Riemann-integrable on $[c - \delta, b]$. Since $f$ is Riemann-integrable on $[c - \delta, b]$, by Theorem 6.1, there exists a partition $P_2$ of $[c - \delta, b]$ such that $U(P_2, f) - L(P_2, f) < \varepsilon/2$. Now consider the common refinement $P = P_1 \cup P_2$ and observe that $U(P, f) - L(P, f) \leq U(P_1, f) + U(P_2, f) - L(P_1, f) - L(P_2, f) < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$. Hence $f$ is Riemann integrable on $[a, b]$ containing $n + 1$ discontinuities. Thus by the principle of mathematical induction we have shown that if $f$ has finitely many discontinuities $n$ in $[a, b]$ then $f$ is Riemann-integrable on $[a, b]$.
Best Answer
$f:[a,c]\to \mathbb{R}$ is Riemann-integrable iff for every $\epsilon > 0$, there is a partition $\mathcal{P}$ of $[a,c]$ such that $$ U(f,\mathcal{P}) - L(f,\mathcal{P}) < \epsilon $$ where $U$ and $L$ denote the upper and lower sums respectively.
Now, if $f\in \mathcal{R}[a,b]$, then there is a partition $\mathcal{P}_1$ of $[a,b]$ such that $$ U(f,\mathcal{P}_1) - L(f,\mathcal{P}_1) < \epsilon $$ and a partition $\mathcal{P}_2$ of $[b,c]$ such that the similar inequality holds.
Now take $\mathcal{P} = \mathcal{P}_1\cup \mathcal{P}_2$, which is a partition of $[a,c]$. Now what can you say about $$ U(f,\mathcal{P}) - L(f,\mathcal{P})? $$ You need to write down the definition of each to find out how this relates to the inequalities involving the $\mathcal{P}_i$'s.