= Solution
Give each nearest-neighbor <edge> of the <cubic lattice> an independent <Bernoulli distribution> state, open with <probability> $p$. The resulting <product measure> is denoted $\mathbb P_p$. In this <bond percolation> model, $C(0)$ is the <percolation cluster> of the origin, and
$$
\theta_d(p)=\mathbb P_p(|C(0)|=\infty),\qquad
\boxed{p_c(d)=\inf\{p\in[0,1]:\theta_d(p)>0\}.}
$$
The uniform-label <monotone coupling of Bernoulli percolation> shows that $\theta_d$ is increasing.
Let $c_n(d)$ count $n$-step <self-avoiding walks> starting at the origin, with $c_0(d)=1$. Splitting a walk after $n$ steps, and discarding the avoidance constraint between the two pieces, gives $c_{n+m}(d)\leq c_n(d)c_m(d)$. The <Fekete lemma> therefore gives the <connective constant>
$$
\boxed{\mu(d)=\lim_{n\to\infty}c_n(d)^{1/n}
=\inf_{n\geq1}c_n(d)^{1/n}.}
$$
The <locally finite graph> structure means that an infinite <percolation cluster> at the origin supplies an open <self-avoiding walk> of every length. Each specified walk has $n$ distinct <edges> and is open with <probability> $p^n$. The <union bound> gives
$$
\theta_d(p)\leq c_n(d)p^n.
$$
For $p\mu(d)<1$ the right side tends to zero. Hence the <connective-constant lower bound for percolation> is $p_c(d)\geq\mu(d)^{-1}$.
For the upper bound first work on the <square lattice>. A finite open <percolation cluster> has an outer boundary containing a simple closed <graph cycle> of dual <edges>, all crossing closed primal <edges>. Such a <dual bond percolation> circuit separates that cluster from infinity. Write $N_\ell$ for the number of simple dual circuits of length $\ell$ surrounding the origin. Each circuit meets the positive horizontal ray at distance at most $\ell$: its bounding box contains the origin and its diameter is bounded by its length. Choose such an intersection as an anchor and orient the circuit. Removing its last edge leaves a rooted <self-avoiding walk> of length $\ell-1$ in the translated <square lattice>. Consequently, for an absolute constant $K$,
$$
N_\ell\leq K\ell c_{\ell-1}(2).
$$
The exact constant and this possible overcount do not matter. If $(1-p)\mu(2)<1$, the <root test> gives
$$
\sum_{\ell\geq L}N_\ell(1-p)^\ell\longrightarrow0
\quad\text{as }L\longrightarrow\infty.
$$
A summable circuit count alone need not give a total sum below one. To use its tail correctly, let $B_R=[-R,R]^2\cap\mathbb Z^2$ and condition every <edge> internal to $B_R$ to be open. This finite event has positive <probability>. A closed dual circuit surrounding all of $B_R$ crosses no internal <edge> of $B_R$, and so its closed-edge <probability> remains $(1-p)^\ell$ under this conditioning. Its length tends to infinity with $R$. Choose $R$ so large that the <union bound> for all such circuits is below one. With positive conditional <probability>, none occurs.
On that event, $C(0)$ contains $B_R$ and cannot be finite: a finite cluster containing $B_R$ would have an enclosing closed dual circuit. Thus $\theta_2(p)>0$ whenever $p>1-\mu(2)^{-1}$. This is the <connective-constant Peierls bound>, proved by excluding short circuits through the open-box conditioning. An embedded coordinate plane in the <cubic lattice> has exactly the same <bond percolation> law as the <square lattice>, so $p_c(d)\leq p_c(2)$. Together,
$$
\boxed{\frac1{\mu(d)}\leq p_c(d)\leq1-\frac1{\mu(2)}.}
$$
There are $2d$ choices for the first step of a <self-avoiding walk> and at most $2d-1$ thereafter, because immediate reversal is forbidden. Hence $c_n(d)\leq2d(2d-1)^{n-1}$ and $\mu(d)\leq2d-1$. In particular $\mu(2)\leq3$. Substituting with the correct directions of the inequalities gives
$$
\boxed{\frac1{2d-1}\leq p_c(d)\leq\frac23.}
$$
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