The Dirichlet function is the pointwise limit of a sequence of Baire class one functions and is itself not Baire class one, so the Baire hierarchy on $[0,1]$ is already strict at the first level

Fix a surjection $s : \mathbb{N} \to \mathbb{Q} \cap [0,1]$ and, for $m \in \mathbb{N}$, let $g_{m} : [0,1] \to \mathbb{R}$ be the indicator of $\{s(0), \dots, s(m)\}$: $g_{m}(x) = 1$ if $x = s(j)$ for some $j \le m$, and $g_{m}(x) = 0$ otherwise.

Suppose, for contradiction, that $D$ is of Baire class one on $[0,1]$.

But $D$ is continuous at no point of $[0,1]$. Let $c \in [0,1]$ and let $\delta > 0$ be real; put $u := \max\{0,\ c - \delta/2\}$ and $v := \min\{1,\ c + \delta/2\}$, so that $u < v$ and $[u,v] \subseteq [0,1] \cap N_{\delta}(c)$, the strict inequality holding because $c \in [0,1]$, $0 < 1$ and $\delta > 0$. The nondegenerate interval $(u,v)$ contains a rational $y_{1}$ and an irrational $y_{2}$, with $D(y_{1}) - D(y_{2}) = 1$, so one of $|D(y_{1}) - D(c)|$ and $|D(y_{2}) - D(c)|$ equals $1$; hence no $\delta$ witnesses continuity at $c$ for $\varepsilon = 1$. This is the argument of the Dirichlet claim, restricted to the domain $[0,1]$.

For $m \in \mathbb{N}$ define $\rho_{m} : [0,1] \to \mathbb{R}$ by $\rho_{m}(x) := \min\{\, |x - s(j)| : j \le m \,\}$, the minimum of a nonempty finite set of reals. Then $\rho_{m}(x) \ge 0$, and $\rho_{m}(x) = 0$ exactly when $x = s(j)$ for some $j \le m$, the minimum being attained.

Then the set of points of $[0,1]$ at which $D$ is continuous is dense in $[0,1]$, since $0 < 1$; in particular it is nonempty.

$\rho_{m}$ is $1$-Lipschitz, hence continuous: choosing $j_{0} \le m$ with $\rho_{m}(x) = |x - s(j_{0})|$ gives $\rho_{m}(y) \le |y - s(j_{0})| \le |y - x| + \rho_{m}(x)$, and exchanging $x$ and $y$ gives $|\rho_{m}(x) - \rho_{m}(y)| \le |x - y|$; so $\delta := \varepsilon$ witnesses continuity at every point.

For $m, n \in \mathbb{N}$ define $h_{m,n} : [0,1] \to \mathbb{R}$ by $h_{m,n}(x) := \max\{\, 0,\ 1 - \iota(n)\,\rho_{m}(x) \,\}$; the index runs over the whole of $\mathbb{N}$, the term at $n = 0$ being the constant $1$ since $\iota(0) = 0$, so that $n \mapsto h_{m,n}$ is a sequence in the sense of Sequences of reals: bounded, eventually, frequently, tails, subsequences. Each $h_{m,n}$ is continuous on $[0,1]$, being the pointwise maximum of the constant $0$ and the continuous function $x \mapsto 1 - \iota(n)\rho_{m}(x)$.

For each fixed $m$ the sequence $n \mapsto h_{m,n}$ converges pointwise on $[0,1]$ to $g_{m}$. If $\rho_{m}(x) = 0$ then $h_{m,n}(x) = \max\{0,1\} = 1 = g_{m}(x)$ for every $n$. If $\rho_{m}(x) > 0$ then, taking a natural $n_{0} \ge 1$ with $\iota(n_{0}) > 1/\rho_{m}(x)$, every $n \ge n_{0}$ has $\iota(n)\rho_{m}(x) \ge \iota(n_{0})\rho_{m}(x) > 1$, so $h_{m,n}(x) = 0 = g_{m}(x)$.

Hence each $g_{m}$ is of Baire class one on $[0,1]$, being the pointwise limit of a sequence of continuous functions.

The sequence $(g_{m})_{m \in \mathbb{N}}$ converges pointwise on $[0,1]$ to $D$. If $x \in \mathbb{Q} \cap [0,1]$ then $x = s(k)$ for some $k$, since $s$ is onto, and $g_{m}(x) = 1 = D(x)$ for every $m \ge k$. If $x \in [0,1]$ is irrational then $x \ne s(j)$ for every $j$, so $g_{m}(x) = 0 = D(x)$ for every $m$. Claim 1 is proved.

Steps 2.2 and 1.3 contradict one another, so the assumption of step 1.2 is false and $D$ is not of Baire class one on $[0,1]$: claim 2 holds, and with step 7.1 the example is complete.