Let $n = \lfloor a^{-1/2} \rfloor$ and notice that $n \ge 2$, since $a \le 1/4$. Choose a large enough integer $k$ to cover $S$ by an $n^k \times n^k$ array $Q$ so that the ratio of the area of $S$ to the area of $Q$ is at most $a$. Subdivide $Q$ into $n^2$ congruent square subarrays $Q'$, and notice that the ratio of the area of $S \cap Q'$ to the area of $Q'$ does not exceed $an^2$. Remove the $Q'$ whose interiors are disjoint from $S$. Continuing, each of the remaining $Q'$ for which $(\text{area}(S \cap Q'))/(\text{area} Q') < a$ is then subdivided into $n^2$ congruent square subarrays, and so on and so forth all the way down, to stop at stage $\max\{j: an^2(k-j) > 1\} \le k-2$ or earlier; this is because at stage $j$, for each square $Q^{(j)}$ in the subdivision, whose interior is not disjoint from $S$, the ratio of the area of $S \cap Q^{(j)}$ to the area of $Q^{(j)}$ is at least $n^{2(j-k)}$, and of these $Q^{(j)}$ only those for which this ratio is less than $a$ are subject to further subdivision.