Nomenclature: Let $P$ be a partition of $[a,b]$ defined by $\{t_0,t_1,\dots,t_n\}$, with $a = t_0 < t_1 < \dots < t_n = b.$ Then we defined the upper and lower sums of $f$ on the partition $P$ as $U(f,P) := (\sup_{[t_{i-1},t_i]}f)(t_i – t_{i-1}), L(f,P) := (\inf_{[t_{i-1},t_i]}f)(t_i – t_{i-1})$, respectively. For brevity, we also define $M_i = \sup_{[t_{i-1},t_i]}f$ and $m_i = \inf_{[t_{i-1},t_i]}f$.
Let $f$ be integrable on $[a,b]$, and let $P = \{t_0,\dots,t_n\}$ be a partition of $[a,b]$ with $U(f,P) – L(f,P) < b – a$. We shall prove that for some $i$ we have $M_i – m_i < 1$.
Proof: Assume, to the contrary, that for every $i$ we have $M_i – m_i \ge 1$. Then, observe that $b – a > U(f,P) – L(f,P) = \sum M_i \Delta t_i – \sum m_i \Delta t_i = \sum (M_i – m_i)\Delta t_i \ge \sum \Delta t_i = b – a$, which is a contradiction.
Next we will show that there are numbers $a_1$ and $b_1$ with $a < a_1 < b_1 < b$ with $\sup f_{[a_1,b_1]} – \inf f_{[a_1,b_1]} < 1$.
Proof: Suppose $i \neq 1,n.$ Then choose $a_1 = t_{i-1}$ and $b_1 = t_i$ for the aforementioned $i$ above and we are done. Now suppose $i = 1$. Choose $b_1 = t_1$ and choose any $a_1$ which satisfies $a < a_1 < b_1$. It is clear that $\sup f_{[a_1,b_1]} \le \sup f_{[a,b_1]}$ and $\inf f_{[a_1,b_1]} \ge \inf f_{[a,b_1]}$ as $[a_1,b_1] \subset [a,b_1]$. Therefore, $1 > \sup f_{[a,b_1]} – \inf f_{[a,b_1]} \ge \sup f_{[a_1,b_1]} – \inf f_{[a_1,b_1]}$. If $i = n$, then choose $a_1 = t_{n-1}$ and let $b_1$ satisfy $a_1 < b_1 < t_n$. Then similarly, $\sup f_{[a_1,b_1]} \le \sup f_{[a_1,b]}$ and $\inf f_{[a_1,b_1]} \ge \inf f_{[a_1,b]}$, so $1 > \sup f_{[a_1,b]} – \inf f_{[a_1,b]} \ge \sup f_{[a_1,b_1]} – \inf f_{[a_1,b_1]}$, and so we are done.
Now define the interval $I_i := [a_i,b_i]$. We will prove that there are numbers $a_2, b_2$ with $a_1 < a_2 < b_2 < b_1$ with $\sup f_{I_2} – \inf f_{I_2} < \frac{1}{2}$.
Proof: Suppose, on the contrary, that for any $a_2,b_2$ satisfying $a_1 < a_2 < b_2 < b_1$, we have that $\sup f_{I_2} – \inf f_{I_2} \ge \frac{1}{2}$. Define the intervals let $X$ be the set of of $x$ satisfying $a_1 < x < \frac{a_1+b_1}{2}$ and let $Y$ be the set of $y$ satisfying $\frac{a_1+b_1}{2} \le y < b_1$. Choose $a_2, b_2 \in X$ with $a_2 < b_2$. Then $\sup f_X – \inf f_X \ge \frac{1}{2}$. Similarly, choose $a_2', b_2' \in Y$ with $a_2' < b_2'$. Then $\sup f_Y – \inf f_Y \ge \frac{1}{2}$. Since $X,Y$ are nonempty, we must have that $\sup I_1 = \sup(X+Y) = \sup X + \sup Y$ and similarly for $\inf$. Therefore, $1 > \sup f_{I_1} – \inf f_{I_1} \ge (\sup f_X + \sup f_Y) – (\inf f_X + \inf f_Y) \ge 1$, which is a contradiction.
We may repeat this argument by dividing the interval $(a_1,b_1)$ in two intervals (like above), then we divide the interval $(a_2,b_2)$ into three intervals, and so on until we divide $(a_n,b_n)$ into $n+1$ intervals. This process would look like so: let $A_1$ be the set of $x$ which satisfy $a_1 < x < \frac{b_{n} – a_{n}}{n} + a_{n}$, let $A_2$ be the set of $x$ which satisfy $\frac{b_{n} – a_{n}}{n} + a_{n} \le x < 2\frac{b_{n} – a_{n}}{n} + a_{n}$, and so on until $A_{n+1}$: $(n-1)\cdot\frac{b_{n} – a_{n}}{n} + a_{n} \le x < b_{n}$. We then choose $A_1 \ni a_{n1} \le b_{n1} \in A_1, A_2 \ni a_{n2} \le b_{n2} \in A_2$ and so on until $A_{n+1} \ni a_{n(n+1)} \le b_{n(n+1} \in A_{n+1}$. Now assume that $\sup f_{I_n} – \inf f_{I_n} \ge \frac{1}{n}$, and so $1 > \sup f_{I_1} – \inf f_{I_1} = \sum \sup f_{A_i} – \sum \inf f_{A_i} = \frac{1}{n} \cdot n = 1$, which is a contradiction.
Since $I_i = [a_i, b_i]$ and $a_n \le a_{n+1} \le b_{n+1} \le b_n$, we satisfy the requirements of the Nested Interval Theorem, which implies that there is an $x_0 \in I_k$ for any $k \in \mathbb{N}.$
Now set $\delta = \min(|a_k – x_0|,|b_k – x_0|)$, and let $\varepsilon > 0$ be given, and choose $k \in \mathbb{N}$ such that $1/k \le \varepsilon.$ Then notice that whenever $0 < |h| < \delta$, we have $|f(x_0 + h) – f(x_0)| \le |\sup_{I_k} f – \inf_{I_k} f| < \frac{1}{k} \le \varepsilon$, since $x_0 + h$ is in an open interval containing $x_0$. In other words, $f$ is continuous at $x_0$.
To prove that $f$ is therefore integrable on infinitely many points, we note that $f$ being integrable on $[a,b]$ implies $f$ is integrable on $a \le c \le d \le b$. So by “pretending” that $a = t_0, b = t_0 + \varepsilon$, we may partition $[t_0,t_0 + \varepsilon]$ even further into $\{s_0,s_1,\dots,s_m\}$ and reapply our proof. We continue to do this for any subinterval, and as infinitely many subintervals exist, there will exist infinitely many points.
Best Answer
The fallacy lies in
By construction, $I_1 = X \cup Y$, so $\sup(I_1) = \sup(X \cup Y) = \max\{\sup(X), \sup(Y)\}$. However, we can save this proof by simply reusing the first lemma to find subintervals.
A cleaned-up proof is shown as follows, using same the definitions and notations as the OP used. The OP has clarified that the question refers to Darboux integration (which is in fact equivalent to Riemann integration).
Lemma 1. If the real function $f$ is Darboux integrable on the closed interval $[a, b]$ with $a < b$, and $\epsilon$ is a positive number, then there is a closed interval $I \subseteq (a, b)$ such that $$ \sup f_I - \inf f_I < \epsilon. $$
Proof. By the definition of Darboux integration, let $P = \langle t_0, t_1, t_2, \dotsc, t_n \rangle$ be a partition of $[a, b]$ such that $$ U(f, P) - L(f, P) < \epsilon (b - a). $$ Suppose for the sake of contradiction that $$ \sup f_{I_k} - \inf f_{I_k} \ge \epsilon $$ for all subintervals $I_k = [t_{k-1}, t_k]$, where $1 \le k \le n$. Denote $\Delta t_k = t_k - t_{k-1}$. Then, $$ \begin{split} U(f, P) - L(f, P) &= \sum_{i = 1}^{n} \sup(f_{I_k}) \Delta t_i - \sum_{i = 1}^{n} \inf(f_{I_k}) \Delta t_i \\ &= \sum_{i = 1}^{n} (\sup(f_{I_k}) - \inf(f_{I_k})) \Delta t_i \\ &\ge \sum_{i = 1}^{n} \epsilon \Delta t_i = \epsilon(b - a), \end{split} $$ which contradicts $U(f, P) - L(f, P) < \epsilon (b - a)$. Therefore, there is some (closed) subinterval $I_k$ such that $$ \sup f_{I_k} - \inf f_{I_k} < \epsilon. $$
We are done if $1 < k < n$. If $k = 1$, note that $I = [\frac{t_0 + t_1}{2}, t_1]$ satisfies $$ \sup(f_I) - \inf(f_I) \le \sup(f_{I_1}) - \inf(f_{I_1}) < \epsilon. $$ Similarly, we choose $I = [t_{n-1}, \frac{t_{n-1} + t_n}{2}]$ if $k = n$. In all cases, there is a closed interval $I \subseteq (a, b)$ such that $$ \sup f_I - \inf f_I < \epsilon. $$
Lemma 2. If the real function $f$ is Darboux integrable on the closed interval $[a, b]$ with $a < b$, then $f$ is continuous at some $x_0 \in [a, b]$.
Proof. The previous lemma gives a closed subinterval $I_1 \subseteq [a, b]$ such that $\sup(f_{I_1}) - \inf(f_{I_1}) < 1$. But $f$ is also Darboux integrable on $I_1$, so the same lemma gives a closed subinterval $I_2 \subseteq I_1$ such that $\sup(f_{I_2}) - \inf(f_{I_2}) < \frac{1}{2}$. This recursive process gives a sequence of closed intervals $I_1, I_2, I_3, \dotsc$ such that $$ I_{k+1} \subseteq \operatorname{int} (I_k) $$ and $$ \sup(f_{I_k}) - \inf(f_{I_k}) < \frac{1}{k} $$ for all $k \ge 1$. Therefore, there is some $x_0 \in \mathbb{R}$ such that $x_0 \in \operatorname{int}(I_k)$ for all $k$ (where $\operatorname{int}(I_k)$ denotes the interior of $I_k$, namely $\operatorname{int}([a, b]) = (a, b)$ when $a < b$).
We now show that $f$ is continuous at $x_0$. For any given $\epsilon > 0$, choose $k$ such that $\frac{1}{k} < \epsilon$, and choose $\delta > 0$ such that $(x_0 - \delta, x_0 + \delta) \subseteq I_k$. Then, for all $x \in (x_0 - \delta, x_0 + \delta)$, $$ \lvert f(x) - f(x_0) \rvert \le \sup(f_{I_k}) - \inf(f_{I_k}) < \frac{1}{k} < \epsilon. $$ Therefore, $f$ is continuous at $x_0$.
Corollary 3. If the real function $f$ is Darboux integrable on the closed interval $[a, b]$ with $a < b$, then $f$ is continuous at infinitely many $x \in [a, b]$.
Proof. Let $x_k = a + \frac{1}{k} (b - a)$. Apply the previous lemma to each of the subintervals $[x_2, x_1], [x_3, x_2], [x_4, x_3], \dotsc$ to see that $f$ is continuous at (at least) one point in each of these (infinitely many) subintervals.
Corollary 4 (bonus). If the real function $f$ is Darboux integrable on the closed interval $[a, b]$ with $a < b$, then the set of points at which $f$ is continuous is dense in $[a, b]$.
The proof follows straightforwardly from the definition of a dense set.