Is there a set $A \subset [0,1]$ such that $\int_{A \times A^\text{c}} \frac{\mathrm{d} x \, \mathrm{d} y}{\lvert x - y\vert}=\infty$?

An alternating sequence of $2n$ strips in a width $w$ provides around $w\log n$ to the sum. Set aside $1/n/(\log n)^2$ width for each $n$. They have a finite total width but infinite total contribution.
Edit:
If $a\lt b\le c\lt d$ then $$\int_a^b dx\int_c^d dy\frac1{y-x}=\\(d-a)\log(d-a)-(d-b)\log(d-b)-(c-a)\log(c-a)+(c-b)\log(c-b)$$ and the last term vanishes if $b=c$.
Start with an interval $[0,2n]$, alternating strips of length 1. There are $n^2$ contributions to the integral. In $2n-1$ cases, the strips are adjacent, contributing $(2n-1)(2\log2-2\log1+0$ .
In $2n-3$ cases there is a gap of two strips, contributing $(2n-3)(4\log4-6\log3+2\log2)$.
In $2n-2k-1$ cases there is a gap of $2k$ strips, contributing $$(2n-2k-1)((2k+2)\log(2k+2)-(4k+2)\log(2k+1)+2k\log(2k))\\ \gt(2n-2k-1)/(2k+1)$$ If $k\lt n/2$ this is more than $n/(2k+1)$ so the total is at least $n\log n/2$.
Reduce the width by a factor $2n$ so it fits a total width of $1$, and it contributes at least $(\log n)/4$. Also, if shrunk to a total width $w$, it contributes $w(\log n)/4$. For each $n$, set aside $$w_n=\frac1{n(\log n)^2}$$ This has a finite sum by the integral test. The total contribution to the integral is at least $\sum 1/(4n\log n)$ which has an infinite sum by the integral test.

Let $b_n=\sum_{k=3}^{n-1} w_k$. Then $$A=\cup_{n=3}^\infty\cup_{k=0}^{n-1}\{b_n+\frac{w_n}{2n}[2k,2k+1)\}$$


This is just to show why/how a construction simpler than the one used by @Empy2 does not work.

For $n \geq 1$ , consider the intervals $$ X_n = [ 1/\log(n+1) , 1/\log(n+1/2)) \qquad Y_n = [1/\log(n+1/2) , 1/\log(n)) $$ The intervals are all disjoint. Define $A = X_0 \cup X_1 \cup X_2... $ for an opportune $X_0$, so that $A^c = Y_0 \cup Y_1 \cup Y_2... $ for an opportune $Y_0$. We are not interested in $Y_0$ and $X_0$, only to the asymptotic behavior of the integral over the domain $X_n \times Y_n$ for large $n>0$.

Define $i_n = \int_{X_n \times Y_n} \frac{dx \, dy}{|x-y|}$. The exact result can be calculated for every $n>0$, but the expression is long and not interesting. The interesting thing is that for large $n$ it can be proven that $$ i_n \sim 1/(n \log(n)^2) $$ Therefore the series of the $i_n $ converges and so the integral over $A \times A^c$. You can try to consider $$ X_n = [ 1/\log(\log(n+1)) , 1/\log(\log(n+1/2)) \qquad Y_n = [1/\log(\log(n+1/2)) , 1/\log(\log(n)) ) $$ which makes the convergence of $\sum i_n$ even slower: in this case you have $$ i_n \sim 1/(n \log(n) \log(\log(n))^2) $$ that still gives a (very slowly) converging series. You can add other log to the definition of the intervals, but this makes the convergence slower and slower, never divergent. So the point is to seek for constructions where the ``number of disjoint intervals is bigger''.

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Integration

Lebesgue Measure

Lebesgue Integral

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