= Solution
The previous part and the permitted assumption $p_c(\mathrm{bond})<1$ imply $p_c(\mathrm{site})<1$. We build an independent <oriented site percolation> process whose open paths carry contact-process infections, and whose site <probability> tends to one as $\lambda$ increases.
Use the <graphical representation of the contact process> with recovery rate one and per-neighbour arrow rate $\lambda$. Fix $\delta>0$. On the parity lattice $(x,n)$ with $n\ge0$ and $x+n$ even, declare $(x,n)$ good if there is no recovery mark at $x$ during $[(n-1)\delta,(n+1)\delta]$, and there is at least one arrow from $x$ to each of $x-1,x+1$ during $[n\delta,(n+1)\delta]$. Extend the Poisson clocks to negative times solely to define the root event; the process itself still starts at time zero. Goodness has <probability>
$$
q(\lambda,\delta)=e^{-2\delta}(1-e^{-\lambda\delta})^2.
$$
The independence here is important. At a fixed spatial site $x$, the allowed time indices differ by two. Its two-unit recovery intervals therefore have disjoint interiors, and its outgoing-arrow intervals are disjoint. At different spatial sites, recovery clocks are different and outgoing arrows have different tails, hence use different directed-edge clocks. Every defining random input belongs to just one parity site's event, up to deterministic interval endpoints at which a Poisson mark has <probability> zero. Thus the good-site indicators really are independent, with common <probability> $q$; no theorem about dependent percolation is needed.
Suppose $(x,n)$ and $(x\pm1,n+1)$ are both good and $x$ is infected at time $n\delta$. The source has no recovery before $(n+1)\delta$, and it sends an arrow to the chosen neighbour during that interval. The target's good event excludes recovery throughout $[n\delta,(n+2)\delta]$, so the target is infected at time $(n+1)\delta$. Induction shows that any infinite oriented path of good sites starting at $(0,0)$ sustains infection at every time layer. Along each transmitting interval there is always an infected site, so the <contact process> never reaches its empty absorbing state.
Take $\delta=\lambda^{-1/2}$. Then
$$
q(\lambda,\lambda^{-1/2})=e^{-2/\sqrt\lambda}(1-e^{-\sqrt\lambda})^2\longrightarrow1.
$$
For some finite $\lambda$ it exceeds $p_c(\mathrm{site})$, and the good-site process has positive survival <probability>. The <contact process> begun from the origin therefore also survives with positive <probability>. Hence
$$
\boxed{\lambda_c<\infty.}
$$
This is the <independent oriented-percolation comparison for the contact process>. If the infection convention assigns total rate $\lambda$ rather than rate $\lambda$ to each of the two neighbours, replace the arrow rate by $\lambda/2$; the same limit proves the same finiteness conclusion.
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