As per the OP's request:
LEMMA Let $f$ be Riemann integrable over $[a,b]$. If $P=\{t_0,\dots,t_n\}$ is a partition of $[a,b]$ such that $$U(f,P)-L(f,P)<\frac {b-a}m$$ with $m\in\Bbb N$; then there exists an $i$ such that $M_i-m_i<\frac 1 m$ where $M_i$ and $m_i$ are the supremum and infimum of $f$ over the subinterval $[t_{i-1},t_i]$.
PROOF IF it were the case $M_i-m_i\geq \frac 1 m $ for each $i$, then
$$\begin{align}\frac{b-a}m &=&\frac 1 m \sum (t_i-t_{i-1})\\
&=&\sum \frac 1 m (t_i-t_{i-1})\\
&\leq& \sum (M_i-m_i) (t_i-t_{i-1})\\
&=&U(f,P)-L(f,P)\end{align}$$
which contradicts our hypothesis.
THEOREM Let $f$ be Riemann integrable over $[a,b]$. Define $$\Gamma =\{x\in[a,b]:f \text{ is continuous in } x\}$$
Then $\Gamma$ is infinite, and dense over $[a,b]$.
PROOF Let $P$ be a partition of $[a,b]$ such that $U-L<1$, where $U$ and $L$ are the upper and lower Darboux sums of $f$. By the previous lemma, there is an $i$ for which $M_i-m_i<1$. If $i\neq 1$ or $i\neq n$, take $a_1=t_{i-1}$,$b_1=t_i$. Else, if $i=1$ take $a_1\in(a,t_1)$ and if $i=n$ take $b_1\in (t_{n-1},b)$. It will be the case that
$$\sup_{A'}f\geq \sup_A f$$
where $A'=[t_0,t_1]$ (resp. $[t_{n-1},t_n]$) and $A=[a_1,t_1]$ (resp. $[t_{n-1},b_1]$) and also that
$$\inf_{B'} f\leq \inf_B f$$
where $B'=[t_0,t_1]$ (resp. $[t_{n-1},t_n]$ and $B=[a_1,t_1]$ (resp. $[t_{n-1},b_1]$. In any case, $a<a_1<b_1<b$ and $$\sup_{I_1}-\inf_{I_1}<1$$ where $I_1=[a_1,b_1]$.
Since $f$ is integrable over $[a,b]$, it is integrable over $[a_1,b_1]$. The very same reasoning but with $m=2$ gives us $a_2,b_2$ with $a<a_1<a_2<b_2<b_1<b$ and with $$\sup_{I_2}f-\inf_{I_2}f<\frac 1 2$$
Continuing the process gives us a sequence of nested closed intervals $\{I_n:n\in\Bbb N\}$ such that $$\sup_{I_n}f-\inf_{I_n}f<\frac 1 n$$
By Cantor's theorem, $\bigcap_{n\in\Bbb N} I_n\neq \varnothing$. Let then $\mu \in \bigcap I_n$. Since the sequences of $b_i$ and $a_i$ are stricticly monotonic, it follows that $\mu$ is none of the extrema. Let $\epsilon >0$ be given, and choose $n\in \Bbb N$ such that $\frac 1 n < \epsilon$. We have that $\mu \in (a_n,b_n)$. Let $\delta >0$ such that if $|\mu-x|<\delta$, $x\in I_n$. Since $\inf f \leq f \leq \sup f$, for each $x\in I_n$, it follows that if $|x-\mu|<\delta$ then $|f(x)-f(\mu)|<\frac 1 n<\epsilon$, and thus $f$ is continuous at $x=\mu$. Since $f$ is integrable over $[a,b]$ it also is integrable over $[a,\mu]$ and $[\mu,b]$. Set $\mu=\mu_1$. By the previous reasoning, we obtain two new different points of continuity, $\mu_2$ and $\mu_3$. Having repeated this $j$ times we would obtain $2^{j+1}-1$ continuity points. Thus $f$ cannot be continuous over a finite number of points $M$, for it would mean there exists $M$ such that $2^{j+1}-1<M$ for each $j\in \Bbb N$. It follows $\Gamma$ is infinite.
Let $x\in[a,b]$ and consider the sequence
$$\sigma_0=[a,x]$$ $$\sigma_n=\left[x-\frac{x+a}{2^n},x\right]$$
The previous argument means there exists for each $n$ a $\gamma_n\in\sigma_n$ with $\gamma_n\in\Gamma$. But then given $\epsilon>0$ we can take obtain an $N$ large enough such that $\frac{|x+a|}{2^n}<\epsilon$ whenever $n\geq N$ and thus $|x-\gamma_n|\leq \frac{|x+a|}{2^n}<\epsilon$ for $n\geq N$, that is, $\gamma_n\to x$. Since $x$ was arbitrarily chosen, $\Gamma$ is dense over $[a,b]$. This completes the proof.
