Given a transient graph G, we define the wired spanning tree as the measure on G that arises when we take an arbitrary monotone seqeuence of finite subgraphs $G_n \subset G$ with $G_n \nearrow G$ and consider $\hat{G_n} = G_n \cup \{w_n\}$ where $w_n$ is the identification of the complement $G_n^c$ and there is an edge to it if there is an outgoing edge of any vertex in $G_n$. Then the wired spanning tree is the weak limit of spanning tree measures on $\hat{G_n}$.
Now I would like to show the following ($T^w$ denotes the wired spanning tree):
If for all $x \in V$ $\inf \limits_{y \in V} P(\text{x,y are connected in } T^w) = 0$, then
$T^w$ consists of infinitely many trees.
My curent approach is to show that one can find for all $\epsilon > 0, K \in \mathbb{N}$ $v_1,...,v_K$ s.t.
$$
P(v_i,v_j\text{ are connected in } T^w) < \epsilon \; \forall i < j
$$
I could already show that there are indeed infinitely many $y \in V$ satisfying $P(\text{x,y are connected in } T^w) < \epsilon$ for a fixed x. Then I would like to go on by induction and for that show that if $v_1,...,v_K$ have already been chosen, I can find a $v_{K+1}$ that also satisfies the inequality.
Then I could show the statement immediately.