Infinite version of fox in a hole

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There is a hole numbered $n$ for every $n\in\mathbb{N}$, and a fox is in one of them. Every morning you can check a hole and win if you catch it there. Otherwise that evening the fox moves from its hole $k$ to one of $\left\{k-1,k+1\right\}\cap\mathbb{N}$. The process repeats indefinitely until the fox is caught. Do you have a strategy to catch it eventually, regardless of how it moves?

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Solution:

Lemma: if at some point the fox is at hole $k_1$ when we check hole $k_2$ where $k_1\le k_2$ have the same parity, then we will catch it in at most $k_2-2$ days if we check holes $k_2-1,k_2-2,\dots,2$ on subsequent days.

The overall strategy is this. For $n=1,2,\dots$, round $n$ consists of checking hole $2^{n+1}$ on day $$2^1-1+2^2-1+\dots+2^n-1=2^{n+1}-2-n,$$ and decreasing the hole index every following day until hole $2$ has been checked. Clearly for some $n$ the sufficient condition of the above lemma holds, so the fox will be caught on round $n$.