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Take the increasing event that a finite square torus graph contains some long horizontal or vertical rectangle crossing. Translations and a quarter turn make this a transitive increasing event on its bonds. A positive critical crossing bound and the Friedgut-Kalai sharp threshold theorem make its supercritical probability tend to one. A bounded family of shorter, thicker rectangles intercepts every possible long crossing. The square-root trick for positively associated events transfers the high probability to each fixed intercepting rectangle. Using every possible translate would lose the useful bounded exponent.

Let $T$ be the torus graph $(\mathbb{Z}/(5n)\mathbb{Z}{)}^2$, with independent bond density $p$. For $n$ large, all rectangles below are too small to identify any of their own vertices or edges. Let $E_n$ be the event that somewhere in $T$ there is a horizontal open crossing of a $4n$ by $n$ rectangle or a vertical open crossing of an $n$ by $4n$ rectangle. It is increasing and invariant under translations and a quarter turn. These symmetries act transitively on all $50n^2$ bonds, including both orientations. Thus it is a transitive increasing event.

A fixed $4n$ by $n$ rectangle has its ordinary planar distribution, so the supplied lower bound gives ${\mathbb{P}}_{1/2}^T(E_n)>c_4$. Choose any $0<\eta <\min (c_4,1/2)$. Since $p-1/2>0$ is fixed and $\log (50n^2)\to \infty$, the Friedgut-Kalai sharp threshold theorem implies ${\mathbb{P}}_p^T(E_n)>1-\eta$ for all sufficiently large $n$. As $\eta$ is arbitrary,It remains to transfer this symmetric statement to a fixed rectangle, without a growing union-bound loss. On each circular coordinate take the ten positions $a_j=\lfloor jn/2\rfloor$, $0\le j<10$. Consecutive circular gaps are at most $\lceil n/2\rceil$. Use the one hundred $3n$ by $2n$ rectangles starting at pairs $(a_i,a_j)$ and their one hundred quarter-turn images. Every crossed $4n$ by $n$ rectangle contains the full horizontal width of one of the first family, with its vertical span contained in that rectangle's height-$2n$ span: the permissible starting intervals in each coordinate have length $n$, so the chosen positions always supply a suitable start. The crossing, restricted between those two vertical sides, gives a crossing of the chosen $3n$ by $2n$ rectangle. The analogous assertion holds in the rotated family.

Consequently $E_n$ implies at least one of these two hundred increasing crossing events. Their common success probability is $a_n=h_p(3n,2n)$. The nth root trick givesOnly two hundred translates are used, independently of scale. This bounded exponent is the essential finite-torus crossing amplification step.

For a requested height $N$, set $s=\lfloor N/2\rfloor$. Glue $k$ width-$3s$, height-$2s$ rectangles, each successive pair overlapping in a square of side $2s$. Their horizontal crossings and the $k-1$ vertical overlap crossings imply a crossing of the union, which has width $(k+2)s$. An overlap square has crossing probability at least $a_s=h_p(3s,2s)$. Thus the Harris lemma givesChoose $k$ fixed and large enough that $(k+2)s\ge \lambda N$ for all sufficiently large $N$. Increasing height and shortening width can only help, so uniformly for $m\le \lambda N$,In particular it exceeds $1-\epsilon$ for every sufficiently large $N$, with the bound depending only on $p,\lambda ,\epsilon$.