Baire's theorem: a Baire class one function on a closed bounded interval $[a,b]$ is continuous at the points of a dense subset of $[a,b]$ that is the trace of a $G_\delta$ set, so its set of discontinuities is meager

Refinement claim. Let $[c,d] \subseteq [a,b]$ with $c < d$ and let $\varepsilon > 0$ be real. For $N \in \mathbb{N}$ put $E_{N} := \{\, x \in [c,d] : |f_{n}(x) - f_{m}(x)| \le \varepsilon/4 \ \text{for all } n, m \ge N \,\}$.

Claim 1. Fix a real $\varepsilon > 0$ and let $D_{\varepsilon} := \{x \in [a,b] : \omega_{f}(x) \ge \varepsilon\}$. It is closed in $\mathbb{R}$, being $[a,b] \cap G$ with $G$ closed and $[a,b]$ closed.

Claim 4. The set of continuity points of $f$ on the domain $[a,b]$ is $[a,b] \cap V$ for a $G_\delta$ subset $V \subseteq \mathbb{R}$.

Claim 3. Let $x \in [a,b]$ and let $\rho > 0$ be real. The set $[a,b] \cap N_{\rho}(x)$ contains a nondegenerate closed interval $[c,d]$ with $c < d$, because $a < b$: taking $c := \max\{a,\ x - \rho/2\}$ and $d := \min\{b,\ x + \rho/2\}$ gives $[c,d] \subseteq [a,b] \cap N_{\rho}(x)$ and $c < d$: if $x = a$ then $c = a$ and $d = \min\{b, a + \rho/2\} > a$; if $x = b$ then $d = b$ and $c = \max\{a, b - \rho/2\} < b$; and if $a < x < b$ then $c < x < d$.

Each $E_{N}$ is closed: for fixed $n, m$ the set $\{x \in [c,d] : |f_{n}(x) - f_{m}(x)| \le \varepsilon/4\}$ is the preimage under the continuous function $|f_{n} - f_{m}|$ of the closed set $\{y : y \le \varepsilon/4\}$, hence of the form $G \cap [c,d]$ with $G$ closed, hence closed since $[c,d]$ is; and $E_{N}$ is the intersection of that nonempty family of closed sets over the pairs $n, m \ge N$.

$[c,d] = \bigcup_{N \in \mathbb{N}} E_{N}$: for $x \in [c,d]$ the sequence $(f_{k}(x))$ converges to $f(x)$, so there is $N$ with $|f_{k}(x) - f(x)| < \varepsilon/8$ for all $k \ge N$, and then $|f_{n}(x) - f_{m}(x)| \le |f_{n}(x) - f(x)| + |f(x) - f_{m}(x)| < \varepsilon/4$ for all $n, m \ge N$.

By the interval form of Baire category applied to $[c,d]$ and the sequence $(E_{N})$, there are $N \in \mathbb{N}$ and reals $u' < v'$ with $[u',v'] \subseteq E_{N} \cap [c,d] = E_{N}$.

For every $x \in [u',v']$ one has $|f_{N}(x) - f(x)| \le \varepsilon/4$. Indeed, let $\eta > 0$ be real; since $f_{m}(x) \to f(x)$ there is $m \ge N$ with $|f_{m}(x) - f(x)| < \eta$, and then $|f_{N}(x) - f(x)| \le |f_{N}(x) - f_{m}(x)| + |f_{m}(x) - f(x)| < \varepsilon/4 + \eta$; as $\eta > 0$ was arbitrary this gives $|f_{N}(x) - f(x)| \le \varepsilon/4$.

Put $x_{0} := (u'+v')/2$, so $u' < x_{0} < v'$. Since $f_{N}$ is continuous at $x_{0}$ there is a real $\delta > 0$ with $|f_{N}(x) - f_{N}(x_{0})| < \varepsilon/4$ for every $x \in [a,b]$ with $|x - x_{0}| < \delta$. Put $u := \max\{u',\ x_{0} - \delta/2\}$ and $v := \min\{v',\ x_{0} + \delta/2\}$, so that $u < x_{0} < v$ and $[u,v] \subseteq [u',v']$ with $|x - x_{0}| < \delta$ for every $x \in [u,v]$.

For $x, y \in [u,v]$: $|f(x) - f(y)| \le |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(x_{0})| + |f_{N}(x_{0}) - f_{N}(y)| + |f_{N}(y) - f(y)| \le \varepsilon/4 + \varepsilon/4 + \varepsilon/4 + \varepsilon/4 = \varepsilon$. Hence $\omega_{f}([u,v]) \le \varepsilon$, $\varepsilon$ being an upper bound of the set whose supremum that is.

The refinement claim is proved: for every $[c,d] \subseteq [a,b]$ with $c < d$ and every real $\varepsilon > 0$ there are $u < v$ with $[u,v] \subseteq [c,d]$ and $\omega_{f}([u,v]) \le \varepsilon$. Moreover every $x$ with $u < x < v$ satisfies $\omega_{f}(x) \le \varepsilon$, since $[a,b] \cap N_{\rho}(x) \subseteq [u,v]$ for $\rho := \min\{x-u,\ v-x\} > 0$ and $\omega_{f}$ is monotone under inclusion.

$D_{\varepsilon}$ contains no nondegenerate closed interval. Were $[c,d] \subseteq D_{\varepsilon}$ with $c < d$, the refinement claim applied to $[c,d]$ and to the positive real $\varepsilon/2$ would give $u < v$ with $[u,v] \subseteq [c,d]$ and $\omega_{f}(x) \le \varepsilon/2 < \varepsilon$ for every $x$ with $u < x < v$; such an $x$ lies in $[c,d] \subseteq D_{\varepsilon}$ and so satisfies $\omega_{f}(x) \ge \varepsilon$, which is impossible.

Hence $D_{\varepsilon}$ is nowhere dense: it is closed, so it equals its own closure, and its interior is empty, since an interior point would have a neighbourhood $N_{\rho}(x) \subseteq D_{\varepsilon}$ and then $[x - \rho/2,\ x + \rho/2]$ would be a nondegenerate closed interval inside $D_{\varepsilon}$.

Claim 2. $D = \bigcup_{n \in \mathbb{N}} D_{1/\iota(n+1)}$, and each $D_{1/\iota(n+1)}$ is nowhere dense by step 8.1, so $D$ is a union of a sequence of nowhere dense sets, that is, meager.

Suppose $[c,d] \subseteq D$ with $c < d$ as in step 1.4. Then $[c,d]$ is covered by the sequence $(D_{1/\iota(n+1)})$ of closed sets, so by the interval form of Baire category some $D_{1/\iota(n+1)} \cap [c,d]$ contains a nondegenerate closed interval, contradicting step 7.1. So $[c,d] \not\subseteq D$, and any point of $[c,d] \setminus D$ is a point of $C$ inside $[a,b] \cap N_{\rho}(x)$.

Claims 1, 2, 3 and 4 are therefore proved: claim 1 by steps 1.2, 7.1 and 8.1, claim 2 by step 9.1, claim 3 by steps 1.4 and 10.1, and claim 4 by step 1.3.