= Solution
A subset $A$ of a <metric space> is <nowhere dense set>[nowhere dense] when the interior of its closure is empty. One form of the <Baire category theorem> says that a nonempty <complete metric space> cannot be a <set union>[countable union] of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.
Fix $\varepsilon>0$ and work in the complete interval $[1,2]$. For $N\geq1$ define the closed set
$$
E_N=\{x\in[1,2]: |f(nx)|\leq\varepsilon\text{ for every }n\geq N\}.
$$
The sets are closed by the <continuous function>[continuity] of $f$, and the assumed <pointwise convergence> gives $[1,2]=\bigcup_NE_N$. The <Baire category theorem> therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq[1,2]$ contained in $E_N$.
The intervals $[na,nb]$ overlap once $(n+1)a\leq nb$, which holds for every sufficiently large <natural number> $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\geq N$ and $x\in(a,b)$, and then $|f(t)|=|f(nx)|\leq\varepsilon$. Since $\varepsilon$ was arbitrary, $f(t)\to0$ as $t\to\infty$.
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