= Solution
Fix a root $r$. On a <recurrent graph>, a <simple random walk> started at any vertex hits $r$ <almost surely>. Run <Wilson's algorithm> with root $r$: successively attach the <loop-erased random walk> from each new starting vertex, stopped when it hits the existing tree. Every walk terminates <almost surely>, because that tree already contains $r$.
To compare the two finite approximations, fix a finite set of edges $K$ and put all their endpoints first in the ordering of starting vertices. In the infinite <graph>, these finitely many <Wilson's algorithm> walks have finite lengths <almost surely>. Their visited vertices, together with their neighbors, are therefore contained in $V_k$ for all sufficiently large $k$ on each realization.
Couple the walks in $G[V_k]$, in $G_k^{\mathrm w}$, and in $G$ using the same neighbor choices while they are in the interior of $V_k$. On the event just described, they encounter neither boundary, so the walks, their chronological <loop erasures>, and the resulting partial trees agree. Once every endpoint of $K$ has entered the partial tree, later walks cannot add any edge of $K$: they stop on their first hit of the existing tree. Thus the indicators of membership for all edges of $K$ agree in the free and wired <uniform spanning trees>, with probability tending to one.
The two limiting measures agree on every finite-edge event, and therefore
$$
\boxed{\mathrm{WSF}=\mathrm{FSF}.}
$$
This argument also shows that the common <uniform spanning forest> is a single spanning <tree> on a <recurrent graph>. The equality refers to probability measures, not to independently sampled forests being identical.
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