= Solution
Monotonicity of $f$ gives, for each $n\geq1$,
$$
f(2^{-n})\log2\leq\int_{2^{-n}}^{2^{-(n-1)}}f(t)\frac{dt}{t}.
$$
Hence $\sum_{n\geq1}f(2^{-n})<\infty$; the additional $n=0$ term is finite by continuity at one. Also $f(t)\to0$ as $t\downarrow0$, since a positive limiting lower bound would make the integral diverge.
By <Brownian scaling>, $S_{2^{-n-1}}$ has law $2^{-(n+1)/2}S_1$. Apply the small-value bound for the <Brownian running maximum> to obtain
$$
\begin{aligned}
\mathbb P\left(S_{2^{-n-1}}<2^{-n/2}f(2^{-n})\right)
&=\mathbb P(S_1<\sqrt2 f(2^{-n}))\\
&\leq\frac2{\sqrt\pi}f(2^{-n}).
\end{aligned}
$$
Summing yields \b[the requested finite <probability> sum]. The argument needs no <independence> of these nested-time events.
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