For a random walk in $\mathbb{Z}^d$ with exactly $N$ steps, the probability of any valid path is $(2d)^{-N}$. Now let $A_N$ denote the set of all paths that return to the origin at least once. Why is (for $N\ge 2$) $$\mathbb{P}(A_N)\ge\frac{1}{2d}$$ ?
My first thought was to take the probability of reaching the origin in exactly $2$ steps, which is a subset of $A_N$, however that probability is $(2d)^{-2}$, which I don't think helps.
I don't know of any other bounding "tricks" for probabilities and I can't think of any more explicit way to find this bound.
Your idea was good, you just miscalculated the probability. There are $2d$ nearest neighbours, and the probability to return to the origin via each of them in $2$ steps is $(2d)^{-2}$, so the total probability to return to the origin in $2$ steps is $(2d)^{-1}$.