= Solution
Suppose the <continuous functions> converged pointwise to the rationality indicator. Each $F_N$ is closed, since
$$
F_N=\bigcap_{n\ge N}\left(f_n^{-1}(( -\infty,1/3])\cup f_n^{-1}([2/3,\infty))\right).
$$
<Pointwise convergence> implies that every point belongs to some $F_N$: its eventual values lie near either zero or one. Thus $\mathbb R=\bigcup_NF_N$. By the <Baire category theorem>, some $F_N$ contains a nonempty open interval $I$. Choose a rational $r\in I$ and an irrational $s\in I$. For sufficiently large $n\ge N$, convergence gives $f_n(r)>2/3$ and $f_n(s)<1/3$. The <intermediate value theorem> then forces $f_n$ to take the value $1/2$ between them, at a point in $I\subseteq F_N$. This contradicts the defining condition of $F_N$. Therefore \b[the rationality indicator is not a pointwise limit of <continuous functions>], the <rationality indicator is not Baire class one> result.
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