= Solution
The <Burton-Keane theorem> says that for independent <bond percolation> on the <hypercubic lattice> $\mathbb Z^d$, $d\ge2$, at every fixed $p\in[0,1]$, there is almost surely \b[at most one <infinite percolation cluster>]. It does not assert that such a cluster exists, nor does it assert absence of percolation at the critical parameter. The cases $p=0$ and $p=1$ are immediate; assume $0<p<1$ for the proof.
Let $N_\infty$ be the number of <infinite percolation clusters>. This <random variable> is translation invariant. By <translation ergodicity of Bernoulli percolation>, it is almost surely a constant $k\in\{0,1,2,\ldots,\infty\}$. One way to see ergodicity is to approximate an invariant event by events on finite sets of <edges>, translate one approximation far enough to make the two approximations independent, and then let the approximation error vanish. Its <probability> equals its square, hence is zero or one.
The other probabilistic input is the <finite-energy property of Bernoulli percolation>. Conditionally on all <edges> outside a finite set, every prescribed inside configuration has positive <probability>, namely a product of powers of $p$ and $1-p$. Thus an event of positive <probability> can be followed by a specified finite modification and still gives a positive-probability image event. Outside infinite arms are unaffected.
First exclude a finite $k\ge2$. Some finite box meets at least two <infinite percolation clusters> with positive <probability>. Open all <edges> in and incident to that box. These clusters merge. Opening finitely many <edges> cannot create an <infinite percolation cluster> solely out of finite clusters, since it joins only finitely many such clusters. Hence the number of <infinite percolation clusters> decreases with positive <probability>, contradicting its almost-sure constant value. This rules out every finite multiplicity greater than one.
Now suppose $k=\infty$. With positive <probability> a finite box meets at least three distinct <infinite percolation clusters>. Retain an exterior infinite ray from each of three of them. Replace the finitely many interior and boundary <edges> by a minimal <tree> joining their three attachment <graph vertices>, and close all unused <edges> there. The exterior rays belong to distinct original clusters, so have no outside connections to each other. The resulting joined cluster has a branch <graph vertex> $v$ with exactly three incident open <edges>, and deleting $v$ leaves three infinite components. If a terminal lies in the middle of the interior connecting path, its exterior ray supplies the third branch there. Such a <graph vertex> is a <trifurcation vertex in percolation>. A finite box has only finitely many candidate locations and modifications; therefore some fixed <graph vertex> has positive trifurcation <probability>. By translation invariance, that <probability> is a common number $\tau>0$ at every <graph vertex>.
The decisive deterministic input is the <trifurcation boundary-counting lemma>. Let $T$ be all these degree-three <trifurcation vertices in percolation> in a finite box $B_n$. In each <infinite percolation cluster> meeting $T$, delete the <graph vertices> of $T$, contract each remaining <connected component of a graph> adjacent to $T$ to a node, and retain the incidences with the deleted <graph vertices>, including direct <edges> between <graph vertices> of $T$. This is a finite <tree>: a cycle would connect two neighbours of an original <trifurcation vertex in percolation> without passing through it, contradicting its three-way separation. Every deleted <graph vertex> has degree three. Every leaf component is infinite, since a finite leaf would give a finite branch after removing its adjacent <trifurcation vertex in percolation>. The elementary <tree> degree identity therefore supplies at least $|T\cap C|+2$ infinite leaf components for a cluster containing $T$.
Each such infinite component reaches the outer <graph vertex> boundary $\partial^+B_n$, and different components reach different boundary <graph vertices>. Summing over the clusters yields, in particular,
$$
|T|\le|\partial^+B_n|.
$$
For a nearest-neighbour lattice box,
$$
|B_n|=(2n+1)^d,\qquad
|\partial^+B_n|=2d(2n+1)^{d-1}.
$$
Taking expectations and using the common trifurcation <probability> gives
$$
\tau(2n+1)^d\le2d(2n+1)^{d-1},\qquad
\tau\le\frac{2d}{2n+1}\longrightarrow0.
$$
This contradicts $\tau>0$ and excludes infinitely many <infinite percolation clusters>. Thus only $k=0$ or $k=1$ remains:
$$
\boxed{\mathbb P_p(N_\infty\in\{0,1\})=1\quad\text{for every }p\in[0,1].}
$$
The proof explains the role of the <amenable boundary growth of lattice boxes>: <finite-energy property of Bernoulli percolation> would make branching points have positive volume density, while their disjoint infinite branches can only be supplied through a boundary of lower order. It uses no planar duality and works in every indicated dimension.
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