Proof that stochastic process on infinite graph ends in finite step. Infinite Graph
Let $G$ be an infinite graph that is constructed this way: start with two unconnected nodes $v_1$ and $u_1$. We call this "level 1".
Create two more unconnected nodes $v_2$ and $u_2$. Connect $v_1$ to both of them with directed edges pointing from $v_1$ to $v_2$, and from $v_1$ to $u_2$ respectively. Then, connect $u_1$ to both of them using directed edges in the same way. This is "level 2".
Repeat this process. At level $n$, nodes $v_n$ and $u_n$ are both connected to the each of the next two nodes $v_{n+1}$ and $u_{n+1}$. This results in an infinite connected graph.
Stochastic Process
We define a discrete time process heuristically on this graph. $v_1$ and $u_1$ are initially "infected". This infection only last for one time period. We start at $t = 0$.
At each time step, infected nodes have an independent probability $p$ of passing the infection to their neighbors (infected node must have a directed edge to the target for the target to be a neighbor). If this happens, the neighbor is infected for one time period.
E.g. $v_1$ is connected to $v_2$ and $u_2$. $v_1$ is infected initially. At time $t=0$, $v_1$ has a probability $p$ of infecting its neighbors $v_2$ and $u_2$. Suppose it successfully infects $v_2$, and $u_2$ is never infected by any of its predecessor. At $t=1$, $v_1$ and $u_1$ stops being infectious. And $v_2$ is infectious.
This process is repeated for each level. If a node is infected twice, it is still just "infected", there is no special meaning to a double infection.
Conjecture
This process ends with probability $1$ after a finite number of steps.
How do I prove this? My attempt: The probability that the infection on one level is passed onto the next level is $q = 1 - (1-p)^4 = $ probability that all $4$ attempts to infect fails. For the infection to reach the $k$th level, the probability is $q^k$ and this tends to zero as $k$ tends to infinity?

If you direct the edges then the problem becomes trivial.
Assume inductively that at time $i$ the only nodes that might be infected are $u_i$ and $v_i$.
Then at time $i+1$ nodes $u_{i+1}$ and $v_{i+1}$ may become infected by $u_i$ and $v_i$, but $u_i$ and $v_i$ become cured, and there is no way of reinfecting them from below.
Thus your argument is correct, and the probability that the process is alive at time $i$ is at most $q^i$. 
Now, for any natural number $n$, if the process does not terminate after a finite number of levels, then the process much reach level $n$.
Therefore the probablity that the process does not terminate is less than or equal to the probability that the process reaches level $n$.
So if $A$ is the event that the process survives indefinitely we have
$$0\leq\mathbb P(A) \leq q^n$$for every $n\in\mathbb N$. Therefore we must have $\mathbb P(A) = 0$.
