In proving that a simple symmetric 2-d random walk a.s. returns to the origin, the proofs generally start by showing (*) that the expected number of returns to the origin is infinite, and then use a lemma that:
Lemma 1: If that expected number of returns is infinite, then the probability of return to the origin is $1$.
The demonstration of Lemma 1, for example in Durrett, Probability Theory and Examples p. 164, is what is confusing me. Because whatever proof holds, it has to break down for the following non-symmetric position and path dependent 2-d lattice random walk $W$:
All steps are units in each of the $+x, -x, +y, -y$ directions unless step $1$ has been taken and is $+x$.
If step $1$ has been taken and is $+x$, then all subsequent steps are $+x$.
$W$ clearly has a probability of return to the origin that is $\leq \frac34$. Yet (by the same reasoning that shows (*) for a simple symmetric 2-d walk) the expected number of returns to the origin is infinite. So that seems to violate Lemma 1.
What am I missing here?
Your question is a very natural one, but what you are missing is the (strong) Markov property. Roughly speaking, this says that when the process returns to its starting position, then it starts over, independently of its past.
If $N$ denotes the total number of visits to the initial position (including the time zero visit), then the Markov property guarantees that $N$ has a geometric distribution with parameter $e$, where $e$ the probability of escape, never to return.
In particular, if $e>0$, then $\mathbb{E}(N)={1\over e}<\infty.$
Therefore if $\mathbb{E}(N)=\infty$, we must have $e=0.$