= Solution
For independent <bond percolation> on $\mathbb Z^d$, $d\ge2$, \b[there is almost surely at most one <infinite percolation cluster> at every parameter]. This is the <Burton-Keane theorem>. It does not require an assumption about the behavior at the critical <probability>. If the <probability> that the origin lies in an <infinite percolation cluster> is positive, <translation ergodicity of Bernoulli percolation> also gives almost-sure existence and hence exactly one.
The two structural ingredients are translation invariance and <finite-energy property of Bernoulli percolation>. For $0<p<1$, every prescribed configuration of finitely many bonds has positive conditional <probability> given all the other bonds. In particular, opening or closing a fixed <finite set> of bonds transforms events of positive <probability> into events of positive <probability>. The product measure is ergodic under translations. One way to see this is to approximate a translation-invariant event by a finite-cylinder event, translate the cylinder sufficiently far to make its coordinates disjoint, and use independence. Sending the approximation error to zero gives $\mathbb P(A)=\mathbb P(A)^2$.
Let $K$ be the number of <infinite percolation clusters>. It is translation-invariant, so it is a fixed value in $\{0,1,2,\ldots,\infty\}$ almost surely. First exclude a finite value $k\ge2$. Some finite box meets at least two <infinite percolation clusters> with positive <probability>. Force every internal bond in that box to be open. This joins those clusters and can only merge other clusters; it cannot split an <infinite percolation cluster>. It also cannot create a new <infinite percolation cluster> out of finite ones, since only finitely many clusters touch the modified region. Thus the number of <infinite percolation clusters> decreases. <finite-energy property of Bernoulli percolation> makes the modified event have positive <probability>, contradicting its almost-sure value $k$.
Suppose next that $K=\infty$. Some finite box meets three different <infinite percolation clusters> with positive <probability>. Each such cluster has an infinite component outside the box: deletion of a <finite set> in a locally finite <graph> leaves only finitely many components adjacent to that set, so an <infinite percolation cluster> cannot have all of them finite. Choose one infinite exterior component from each of the three clusters, and one bond connecting it to the box.
Inside the box, connect the three selected exterior endpoints by a finite <tree>, preserving just their three selected entrance bonds. Take the minimal subtree connecting these endpoints; they are leaves, since their only retained connection into the box is their chosen entrance bond. This <tree> has a branch <graph vertex> of degree three. Close every other internal or boundary bond of the box. Its three exterior components remain distinct outside, since they belonged to different original clusters. The branch <graph vertex> now has exactly three open incident bonds, and deleting it separates its cluster into three infinite components. Such a <graph vertex> is a <trifurcation vertex in percolation>. There are finitely many choices for the endpoints and <tree> within the chosen region, so at least one fixed <finite modification of Bernoulli percolation> has positive <probability>. <finite-energy property of Bernoulli percolation> consequently gives positive <probability> of a trifurcation at some fixed <graph vertex>. Translation invariance makes this <probability> the same number $q>0$ at every <graph vertex>.
The decisive fact is that a finite box cannot contain a positive-volume density of these branching points. Here is the boundary count, including its combinatorial justification. For a finite <graph vertex> set $W$, let $\partial^+W$ be the exterior <graph vertices> adjacent to it. In each <infinite percolation cluster> meeting $W$, contract each of its components outside $W$ that is adjacent to $W$ to a terminal <graph vertex>. The quotient <graph> consisting of the cluster's <graph vertices> in $W$ and these terminals is finite and connected. A trifurcation <graph vertex> in $W$ separates its terminals into at least three nonempty groups, since each of its infinite branches must leave $W$.
Take a minimal subtree of the quotient <graph> connecting all the terminals. Every such trifurcation <graph vertex> must belong to that subtree and have degree at least three: otherwise the terminals in one of its separated groups could not connect to the other groups. All leaves of the minimal subtree are terminals. A cluster with only one terminal has no counted trifurcation. For a finite <tree> with at least two <graph vertices> and $L$ leaves,
$$
\sum_v(\deg(v)-2)=-2
$$
implies that the number of <graph vertices> of degree at least three is at most $L-2$, and therefore at most its number of terminals. Distinct exterior components can be assigned distinct <graph vertices> of $\partial^+W$. Terminals from different clusters are disjoint as well. Summing over clusters proves the <trifurcation boundary-counting lemma>
$$
\#\{v\in W:v\text{ is a trifurcation}\}\le|\partial^+W|.
$$
Take $W=[-n,n]^d\cap\mathbb Z^d$. The expected left side is $q|W|$, of order $qn^d$, while the right side is $O_d(n^{d-1})$. Division by $|W|$ and passage to infinity force $q=0$, contradicting the positive <probability> obtained by <finite modification of Bernoulli percolation>. Thus infinitely many <infinite percolation clusters> are impossible. The surface-to-volume comparison is the geometric reason for uniqueness on the lattice; the same argument is tied to amenability on more general <graphs>.
At $p=0$ there is no <infinite percolation cluster>, and at $p=1$ the whole lattice is the unique <infinite percolation cluster>. This completes all parameter cases. Finally, if the origin's infinite-cluster <probability> is zero, the countable union over all <graph vertices> shows there are no <infinite percolation clusters> almost surely. If it is positive, the translation-invariant existence event has positive <probability> and hence <probability> one. Together with uniqueness this distinguishes absence from the case of a single <infinite percolation cluster>, without making an unsupported assertion that every <graph vertex> belongs to it.
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