= Solution
Let $E_T,E_B,E_L,E_R$ be the <increasing events> that an infinite simple open <ray in a graph> starts at the indicated side and immediately leaves the box, with all later <graph vertices> outside it. A corner exit is assigned according to the direction of its outgoing <edge>. The four events have the same <probability> $a_N$ by rotational symmetry. Their union is $D_N$ from the preceding step.
The <Harris-FKG inequality> applies to these exterior-ray events by approximation with finite connections to larger boxes and passage to decreasing limits. Positive association of the four <decreasing events> $E_s^c$, by iterating the <Harris-FKG inequality>, gives
$$
1-\mathbb P(D_N)=\mathbb P\Bigl(\bigcap_s E_s^c\Bigr)
\ge\prod_s\mathbb P(E_s^c)=(1-a_N)^4.
$$
Equivalently, one may apply the preceding <square-root bound for increasing events> first to the two opposite-side unions and then to the two individual sides of one pair. Therefore
$$
a_N\ge1-(1-\mathbb P(D_N))^{1/4}\longrightarrow1.
$$
A <union bound> now shows
$$
\mathbb P(E_T\cap E_B)\ge1-2(1-a_N)
\ge1-2(1-\mathbb P(D_N))^{1/4}.
$$
Choose $N$ such that $1-\mathbb P(D_N)<(1/8)^4$. Then
$$
\boxed{\mathbb P(E_T\cap E_B)>3/4.}
$$
The infinite exterior clusters touched by these two sides need not be the same. The same argument works for dual boxes and any sufficiently large size; this permits the matching primal/dual contours used in the full proof.
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