= Solution
We use <planar duality for rectangle crossings>, <Harris-FKG inequality> and the permitted <exponential decay of subcritical percolation>. A critical-point detail matters: uniqueness stated only for $p>p_c$ cannot by itself be applied at $p=p_c$. We prove the <Burton-Keane theorem> by the <boundary counting proof of percolation uniqueness>, establishing at most one <infinite percolation cluster> at every parameter and avoiding that gap.
For $0<p<1$, the number $N$ of <infinite percolation clusters> is <almost surely> constant by <translation ergodicity of Bernoulli percolation>. It cannot be a finite constant greater than one. A sufficiently large box meets two different <infinite percolation clusters> with positive <probability>; opening its finitely many interior <edges> joins them without creating a new <infinite percolation cluster>, decreasing $N$. <Finite modification of Bernoulli percolation> gives positive <probability> to this modification, contradicting constancy.
Nor can $N=\infty$. A finite box then meets three different <infinite percolation clusters> with positive <probability>. Select one infinite exterior branch from each. In the box and a finite collar retain just an embedded three-armed <tree> joining those branches and close all other incident <edges> there. Its branching <graph vertex> separates three infinite components when removed, so is a <trifurcation vertex in percolation>. All selections concern finitely many possible entrances and <edge> patterns; <finite-energy property of Bernoulli percolation> gives positive <probability> for at least one such pattern. Translation invariance then gives a common positive trifurcation <probability> $q$.
Here is the counting contradiction in detail. For a finite box $W$, contract each open component outside $W$ that touches $W$ to a boundary terminal. In each cluster, the resulting incidence <graph> is finite and connected. Every trifurcation in $W$ separates at least three sets of terminals, since each infinite branch must leave $W$. A minimal subtree joining all terminals must therefore contain that <graph vertex> with <degree of a vertex> at least three. Its leaves are terminals. The <tree> identity $\sum_v(\deg(v)-2)=-2$ bounds the number of those branching <graph vertices> by the terminal count. Different terminals are represented by different exterior <graph neighbours> of $W$. Thus the <trifurcation boundary-counting lemma> gives
$$
q|W|=\mathbb E\bigl[\#\{v\in W:v\text{ trifurcates}\}\bigr]\leq|\partial^+W|.
$$
For $W=[-r,r]^2\cap\mathbb Z^2$, the two sizes are $(2r+1)^2$ and $8r+4$. Letting $r\to\infty$ contradicts $q>0$. Therefore $N\in\{0,1\}$ <almost surely>, with the same conclusion for the translated <planar dual graph>.
Now prove absence of percolation at $p=1/2$ by <Zhang's argument>. Suppose instead that $\theta(1/2)>0$. <Almost surely> an infinite primal cluster exists, and self-duality gives an infinite dual cluster as well. An infinite connected subgraph of this locally finite lattice contains a <graph ray>, by the <König infinity lemma>. Expanding square boxes meet the <infinite percolation clusters> with <probability> tending to one.
For $B_r=[-r,r]^2\cap\mathbb Z^2$, let $A_N,A_E,A_S,A_W$ mean that an open <graph ray> takes an outward <edge> through the indicated side and subsequently uses <graph vertices> outside $B_r$. The union occurs whenever the box meets an <infinite percolation cluster>: take the last exit of an infinite <graph ray> from the finite box. Conversely an exterior arm supplies an <infinite percolation cluster> meeting its boundary. Quarter-turn symmetry makes the four <probabilities> equal. Their complements are <decreasing events>, so the <square-root trick for positively associated events> gives
$$
\mathbb P_{1/2}(A_N)\geq1-\mathbb P_{1/2}(B_r\not\leftrightarrow\infty)^{1/4}\longrightarrow1.
$$
For the <planar dual graph> use the box with boundary coordinates $\pm(r+1/2)$ and define its exterior side-arm events in the same way. This box also has quarter-turn symmetry and meets the dual <infinite percolation cluster> with <probability> tending to one. Thus each dual side has an infinite exterior arm with <probability> tending to one. Primal outward <edges> cross this dual-box contour on the corresponding sides; their remaining <graph vertices> stay outside it. A <union bound> makes the simultaneous event of primal north/south arms and dual east/west arms have positive <probability> for a sufficiently large box.
The four arms alternate around the contour. Open all <edges> with both endpoints in $B_r$, leaving all outward and exterior <edges> intact. This joins the two primal entrance points and preserves all four exterior arms. <Finite-energy property of Bernoulli percolation> keeps the event's <probability> positive. The primal north and south arms are now connected through the box. The two dual arms must lie in different infinite dual clusters. To see the separation, a hypothetical finite dual <graph path> joining the east and west arms cannot enter the dual-box interior: its boundary-crossing and interior <edges> cross primal <edges> that were just opened. Erase its loops and cut it at successive hits of the contour. A resulting exterior dual crosscut, together with the corresponding contour arc, encloses one of the two intervening primal entrance points. The infinite primal <graph ray> from that point would have to cross a dual-open <edge>, which is impossible. This is the planar separation used in the <alternating arms argument at the self-dual percolation parameter>. It contradicts uniqueness of the infinite dual cluster. Hence
$$
\boxed{\theta(1/2)=0,\qquad p_c(2)\geq1/2.}
$$
Finally suppose $p_c(2)>1/2$. At $p=1/2$ the permitted <exponential decay of subcritical percolation> would supply $C,c>0$ with $\mathbb P_{1/2}(v\leftrightarrow\partial B_n(v))\leq Ce^{-cn}$. Use the rectangle with <graph vertices> $\{0,\ldots,n\}\times\{0,\ldots,n-1\}$, and let $H_n$ be its left-to-right open crossing. <Planar duality for rectangle crossings> identifies its complement with a dual top-to-bottom crossing of a rectangle of width $n-1$ and height $n$. Rotation and translation give the same crossing law; side <edges> at the entrance and exit boundaries are irrelevant. Thus the <exact self-dual rectangle crossing probability> is
$$
\mathbb P_{1/2}(H_n)=\frac12.
$$
But a crossing has some starting <graph vertex> on its $n$-vertex left side connected to distance $n$. The <union bound> and <exponential decay of subcritical percolation> imply $\mathbb P_{1/2}(H_n)\leq nCe^{-cn}\to0$, a contradiction. Therefore
$$
\boxed{p_c(2)=1/2\quad\text{and}\quad\theta(1/2)=0.}
$$
No Russo-Seymour-Welsh theorem or continuity assertion for $\theta$ is being used, and critical uniqueness was proved rather than inferred from the supercritical hypothesis.
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