 Solution

A self-avoiding walk is a finite sequence of graph vertices $v_0,\dots ,v_n$ in which successive vertices are adjacent and no graph vertex is repeated; its length is $n$, the number of edges. Include the zero-length walk.

Split a length-$m+n$ walk from $v$ into its first $m$ edges and remaining $n$ edges. After fixing the first segment, the number of possible suffixes is at most ${\sigma }_n$, since forgetting the prohibition on revisiting its earlier vertices can only increase that number. Thus ${\sigma }_{m+n}(v)\le {\sigma }_m(v){\sigma }_n$, and taking the supremum givesEvery finite-radius ball is finite because degrees are bounded. An infinite connected graph has vertices arbitrarily far from each root, so shortest paths give ${\sigma }_n(v)\ge 1$. Also ${\sigma }_n\le \Delta (\Delta -1{)}^{n-1}$ for $n\ge 1$: after the first edge, an immediate reversal is forbidden. In particular $\Delta \ge 2$ and these suprema are finite.

Apply the Fekete lemma: a real subadditive sequence $a_{m+n}\le a_m+a_n$ satisfies $\lim a_n/n={\inf }_{n\ge 1}{a}_n/n$, allowing negative infinity. Here $a_n=\log {\sigma }_n\ge 0$, so the limit is finite. The uniform connective constant of a bounded-degree graph is therefore