= Solution
Use consistently wired boundary conditions in increasing finite boxes and let $\phi_{p,q}^{\mathrm w}$ be the usual wired infinite-volume <random-cluster measure>. The finite-volume comparison just proved applies with boundary <graph vertices> identified, since it is still a comparison on a finite graph. Passing to the infinite-volume limit preserves the inequality for every <increasing event> determined by finitely many <edges>.
For fixed $m$, let $A_m$ be the event that the origin connects by an <open path> in $[-m,m]^2$ to its boundary. It is such a finite <increasing event>. Therefore, for $q_1\le q_2$,
$$
\phi_{p,q_1}^{\mathrm w}(A_m)\ge\phi_{p,q_2}^{\mathrm w}(A_m).
$$
The events decrease to the event that the origin has an <infinite percolation cluster>, by local finiteness. <Continuity from above of a measure> gives
$$
\theta^{\mathrm w}(p,q_1)\ge\theta^{\mathrm w}(p,q_2).
$$
Define the <random-cluster critical probability> by $p_c(q)=\inf\{p:\theta^{\mathrm w}(p,q)>0\}$. The set in which percolation occurs at $q_2$ is a subset of that at $q_1$, whence
$$
\boxed{q_1\le q_2\quad\Longrightarrow\quad p_c(q_1)\le p_c(q_2).}
$$
The same proof works if free boundary conditions are used consistently throughout. It compares critical <probabilities>; it does not presume whether an <infinite percolation cluster> exists at the critical parameter itself.
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