= Solution
Use the <uniform bound for harmonic sine polynomials>
$$
P_m(t)=\sum_{k=1}^m\frac{\sin kt}{k},\qquad
\|P_m\|_\infty\leq C_0,\quad C_0=1+\pi.
$$
For completeness, reduce to $0<t\leq\pi$ and split at $K=\min(m,\lfloor1/t\rfloor)$. The first part is bounded by $Kt\leq1$. Geometric-series summation bounds every interval sum of $\sin kt$ by $1/\sin(t/2)\leq\pi/t$. <Summation by parts> bounds the remaining harmonic-weighted tail by $\pi/[t(K+1)]\leq\pi$. Negative $t$ follows by oddness and $t=0$ is immediate.
Let $H_m=\sum_{k=1}^m1/k$. Choose $m_j$ so large that $H_{m_j}\geq C_0\,2^{j+2}$, and positive integers $L_j>m_j$ such that the intervals $[L_j-m_j,L_j+m_j]$ are strictly separated and increase. Define
$$
Q_j(t)=e^{iL_jt}\frac{P_{m_j}(t)}{H_{m_j}},\qquad
\boxed{g(t)=\frac12\sum_{j=1}^{\infty}Q_j(t).}
$$
The bound $\|Q_j\|_\infty\leq2^{-j-2}$ makes this a continuous function with $g(0)=0$.
Each <Fourier coefficient> of $P_m$ has modulus $1/(2k)$, so the sum of the absolute coefficients is $H_m$. Therefore every prefix of a normalized block $Q_j$ has norm at most one. At any Fourier cutoff, all earlier blocks are complete, at most one block is partial and all later blocks are absent. Hence
$$
\boxed{\|S_ng\|_\infty
\leq\frac12\left(\sum_{j\geq1}2^{-j-2}+1\right)
=\frac58<1.}
$$
At zero, a completed block contributes zero, but its prefix through frequency $L_j$ consists of the negative-frequency sine coefficients and equals $i/2$. Thus
$$
S_{L_j}g(0)=\frac i4,\qquad
S_{L_j+m_j}g(0)=0.
$$
Both index sequences tend to infinity. \b[The uniformly bounded partial sums fail to converge at the origin], even though the function is continuous and zero there.
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