Let $S_n$ be the simple symmetric random walk on $\mathbb Z$. I want to show that the event $(S_n=0,S_{n+1}=1)$ for some $n$] occurs with probability $1$.
I know the SSRW is recurrent so $(S_n=0)$ occurs at least for one $n$ with probability $1$. It seems to be intuitively obvious that the event I wrote also occurs almost surely but I cannot really prove it.
Because the simple random walk is recurrent and has the strong Markov property, we know that the event $S_n=0$ happens for infinitely many $n \in \mathbb{N}$ with probability $1$. If we define a sequence of stopping times $(\tau_k)_{k \in \mathbb{N}}$ by $\tau_0 := 0$,
$$\tau_{k+1} := \inf\{n>\tau_k; S_n(\omega)=0\}, \qquad k \in \mathbb{N}_0,$$
then this implies $\mathbb{P}(\tau_k<\infty)=1$ for all $k \in \mathbb{N}$. For
$$A := \{\exists n \in \mathbb{N}: S_n = 0, S_{n+1}=1\}$$
we have
$$A^c = \bigcap_{k \in \mathbb{N}} \{S_{\tau_0+1}=\ldots = S_{\tau_k+1}=-1\},$$
and so
$$\mathbb{P}(A^c) = \lim_{k \to \infty} \mathbb{P}(S_{\tau_0+1}=\ldots=S_{\tau_k+1}=-1).$$
Since $S_{\tau_k}=0$ for all $k$, this gives
$$\mathbb{P}(A^c) = \lim_{k \to \infty} \mathbb{P}(S_{\tau_0+1}-S_{\tau_0}=-1, \ldots, S_{\tau_k+1}-S_{\tau_k}=-1)$$
and by the independence and stationarity of the increments, we get
$$\mathbb{P}(A^c) = \lim_{k \to \infty}\prod_{j=0}^k \underbrace{\mathbb{P}(S_{\tau_j+1}-S_{\tau_j}=-1)}_{1/2} = \lim_{k \to \infty} \frac{1}{2^{k+1}} = 0.$$