= Solution
We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb T$.
First let $K$ be a finite union of closed arcs with $|K|\le3\varepsilon$, where $\varepsilon$ is small. Choose a <smooth function> nonnegative <function> $u$, equal to one on $K$, with $u\le1$ and $\eta:=\int u\le6\varepsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic <Herglotz integral>
$$
G(z)=\frac{\eta+\displaystyle\int_{\mathbb T}\frac{e^{2\pi is}+z}{e^{2\pi is}-z}u(s)\,ds}{2\eta}.
$$
Its real part is $(\eta+P[u](z))/(2\eta)>0$, where $P[u]$ is the <disk Poisson integral>, and $G(0)=1$. Smoothness of $u$ makes $G$ <continuous> and <smooth function> up to the circle; on $K$ its boundary real part is at least $1/(2\eta)$. The branch $H=\log G$ therefore has $H(0)=0$ and
$$
|\operatorname{Im}H|<\pi/2,\qquad
\operatorname{Re}H\ge\log\frac1{12\varepsilon}\quad\text{on }K.
$$
The <complex logarithm> has no branch problem because $G$ lies in the right half-plane. Its <smooth function> boundary values give absolutely summable Taylor coefficients. Choose an analytic <polynomial> $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\operatorname{Im}A(e^{2\pi it})/\pi$. Then $\|P\|_\infty\le1/2+1/(4\pi)<1$.
If $D=\deg A$ and $m>D$, the modulated <trigonometric polynomial> $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp <Fourier partial sum> at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:
$$
S_m(Q,t)=-\frac{e^{2\pi imt}\overline{A(e^{2\pi it})}}{2\pi i}.
$$
Consequently, for every $t\in K$,
$$
|S_m(Q,t)|\ge B(\varepsilon):=\frac{\log(1/(12\varepsilon))-1/4}{2\pi},\qquad\|Q\|_\infty\le1.
$$
\b[The carrier frequency $m$ can be arbitrarily large.] This <compact-set Fourier amplification lemma> uses a bounded imaginary part to obtain a large one-sided <Fourier multiplier>; no assertion about pointwise divergence is used.
Now let $E$ be any open set of measure $\delta$. Define
$$
\psi(\delta)=\max\{0,B(\delta)/4-1\}.
$$
If this is zero take $f=0$. Otherwise choose a decreasing sequence $\delta_j\downarrow0$, with $\delta_1=\delta$, so small for $j\ge2$ that
$$
2^{-j}B(\delta_j)\ge\psi(\delta)+2.
$$
For $j=1$ this inequality also holds: $B(\delta)>4$ and $B(\delta)/2\ge B(\delta)/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.
Using <compact batching of a small open set>, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3\delta_j$, such that $L_j$ is covered by the interiors of $K_1,\ldots,K_j$ and
$$
\left|E\setminus\operatorname{int}(K_1\cup\cdots\cup K_j)\right|\le\delta_{j+1}.
$$
Here is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most $\delta_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most $\delta_j$ plus an arbitrarily small excess. By <inner regularity of Lebesgue measure>, a further compact subset of the uncovered open portion captures all but $\delta_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3\delta_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset\bigcup_jK_j$.
Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. Set
$$
f(t)=\sum_{j=1}^\infty2^{-j}e^{2\pi im_jt}P_j(t).
$$
The <frequency-separated Fourier block series> and the <Weierstrass M-test> give a <continuous> complex-valued <function> with $\|f\|_\infty\le\sum_j2^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,
$$
|S_{m_j}(f,t)|\ge2^{-j}B(\delta_j)-\left|\sum_{i<j}2^{-i}e^{2\pi im_it}P_i(t)\right|
\ge\psi(\delta)+2-1>\psi(\delta).
$$
Every point of $E$ lies in some $K_j$, so
$$
\boxed{S^*(f,t)\ge\psi(\delta)\quad(t\in E),\qquad
\psi(\delta)\sim\frac1{8\pi}\log\frac1\delta\longrightarrow\infty.}
$$
Values of $\psi$ outside $0<\delta\le1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.
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