= Solution
For $\omega\in\{0,1\}^{E}$, write $o(\omega)=|\eta(\omega)|$ and $c(\omega)=|E|-o(\omega)$. The <random-cluster model> is
$$
\phi_{p,q}(\omega)=\frac1{Z_{p,q}}p^{o(\omega)}(1-p)^{c(\omega)}q^{k(\omega)}.
$$
For $0\leq p\leq1$ and $q\geq1$, its <positive association of the random-cluster model> states that increasing functions $f,g$ satisfy $\phi_{p,q}(fg)\geq\phi_{p,q}(f)\phi_{p,q}(g)$.
For the two parallel edges $e_1,e_2$, the four unnormalized weights for $00,10,01,11$ are respectively
$$
(1-p)^2q^2,quad p(1-p)q,quad p(1-p)q,quad p^2q.
$$
For the increasing events $A=\{e_1\text{ open}\}$ and $B=\{e_2\text{ open}\}$, positive association is equivalent to $w_{00}w_{11}\geq w_{10}w_{01}$, which reduces to $q\geq1$. It therefore fails whenever $p,q\in(0,1)$.
At $p=1/2$, all factors involving the number of open edges are equal, so the weight is proportional to $q^{k(\omega)}$. The smallest possible component count is one. Dividing numerator and denominator by $q$ and sending $q\downarrow0$ leaves equal weight precisely on connected spanning subgraphs, proving the stated <uniform connected-subgraph limit of the random-cluster model>.
Finally let $p,q\downarrow0$ with $q/p\to0$, and let $N=|V|$. Fix a spanning tree $\tau$. The ratio of the weight of $\omega$ to that of $\tau$ is
$$
\left(\frac{p}{1-p}\right)^{o(\omega)-N+1}q^{k(\omega)-1}
=(1-p)^{N-1-o(\omega)}p^{d(\omega)}\left(\frac qp\right)^{k(\omega)-1},
$$
where $d(\omega)=o(\omega)+k(\omega)-N\geq0$. Equality $d=0$ means that the open graph is a forest. The ratio tends to zero unless $d=0$ and $k=1$, which means precisely that $\omega$ is a spanning tree. All spanning trees have equal weight, so the limiting law is the <uniform spanning-tree limit of the random-cluster model>:
$$
\phi_{p,q}(\omega)\longrightarrow
\begin{cases}
1/|\mathcal T|,&\omega\in\mathcal T,\\
0,&\omega\notin\mathcal T,
\end{cases}
$$
where $\mathcal T$ is the set of spanning trees of $G$.
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