= Solution
The <Baire category theorem> says that in a nonempty <complete metric space>, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty <open set> $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline B(x_n,r_n)$ with $0<r_n\leq2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a <Cauchy sequence>; <completeness> supplies a <limit> lying in every closed ball, hence in $O\cap\bigcap_nG_n$. The proof also works at isolated points, where sufficiently small balls are singletons.
The incomplete <metric space> $\mathbb Q$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.
For the sequence in the problem, put
$$
E_N=\bigcap_{n,m\geq N}\{x\in[0,1]:|f_n(x)-f_m(x)|\leq\epsilon\}.
$$
Each $E_N$ is closed by <continuity>. <Pointwise convergence> makes each point's values a <Cauchy sequence>, so $[0,1]=\bigcup_NE_N$. By the <Baire category theorem>, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.
If <continuous> $g_n$ converged pointwise to the rationality indicator, take $\epsilon=1/4$ and let $m\to\infty$ in this bound. On $J$, $|g_n-g|\leq1/4$ for $n\geq N$. Hence a fixed <continuous> $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts <continuity>. Thus \b[the rationality indicator is not a pointwise <limit> of <continuous> functions].
Nevertheless $h_n(x)=x^n$ are <continuous> and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This <limit> is discontinuous at $1$: \b[a pointwise <limit> of <continuous> functions can be discontinuous].
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