I asked a colleague in Hungary, and he found the solution here (on page 170):
http://real-j.mtak.hu/9393/1/MTA_MatematikaiLapok_1992.pdf
It is in Hungarian, but with some effort and Google translator, finally I understood.
EDIT This is the translation of the argument above. I keep the same notation but write $E^c$ for $[0,1] \setminus E$ insetad of $\overline E$.
a) Let $$F=\{t \in E^c: |I \cap E| \leq K|E||I|\}$$ for every interval $t \in I$. In other words, $F$ consists of all points for which the maximal function of $\chi_E$ is less than $K|E|$. Using the maximal inequality, one selects $K$ (independent of $E$) such that $|F| \approx 1$, hence $|E_1|:=|E^c \cap F| \geq 1/2$.
b) Given $s \in E$ let $I_s$ be an interval containing $s$ such that $|I_s \cap E|=(1/2)|I_s|$. Such a $I_s$ exists in all Lebesgue points of $E$, again using that $|E| <1/2$ and a continuity argument.
c) If $s \in E$ and $t \in E_1$, then applying a) to $J=I_s \cup [s,t]$ (or the other way around) we get $(1/2)|I_s|=|I_s \cap E| \leq |J\cap E| \leq K|E|(|t-s|+|I_s|)$. Then $|t-s| \geq \frac{|I_s|}{4K|E|}$ if $K|E| \leq 1/4$.
d) Suppose that $K|E| \leq 1/4$. Then $$ \int_{E_1} \frac{dt}{(s-t)^2} \leq 2 \int_{\frac{|I_s|}{4K|E|}}^\infty \frac{du}{u^2}=\frac{8K|E|}{|I_s|}$$ but also, using b), $$h(s)=\int_{E^c} \frac{dt}{(s-t)^2}\geq \int_{I_s \cap \geq E^c} \frac{dt} {(s-t)^2} \geq \frac{1}{|I_s|^2} \frac 12 |I_s|=\frac{1}{2|I_s|}.$$ Summing up, $$ \int_{E_1} \frac{dt}{(s-t)^2} \leq 16 h(s) K|E|.$$
e) If $K|E| \geq 1/4$ then $$ \int_{E_1} \frac{dt}{(s-t)^2} \leq \int_{E^c} \frac{dt}{(s-t)^2} =h(s) \leq 4h(s) K|E| \leq 16 h(s) K|E|.$$
f) Finally, $$ \int_{E_1}dt\int_{E}\frac{ds }{h(s)(s-t)^2}=\int_{E}\frac{ds}{h(s)}\int_{E_1}\frac{ dt}{(s-t)^2} \leq 16K|E|^2 $$ and there is a point $t \in E_1$ for which the statement holds with $c=32 K$ since $|E_1| \geq 1/2$.
PS This does not deserve badges for maths...maybe for translating from hungarian!
