The strong law of large numbers says : Let $X_1, X_2, \dots$ be i.i.d. integrable random variables defined on $(\Omega, F, \Bbb P)$. Then we have that $$\Bbb P( \frac{\sum_{i=1}^nX_i}{n} \to EX_1) = 1$$
Show thanks to it, that if $EX_1 \neq 0$, the probability that the random walk with steps $X_1, X_2, \dots$ visits the starting point $x$ infinitely often is zero.
So I define $S_k:=\sum_{i=1}^k X_i$ and $V:=\sum_{k=0}^{\infty} \textbf{1}_{\{S_k=x\}}$. We want to show that $\Bbb P(V=\infty)=0$. I wanted to apply SLLN with V not with $S_k$ but the assumption $EX_1 \neq 0$ forces me to use $S_k$ in SLLN.
Hence, what I thought of doing is applying SLLN with the latter, giving $$\Bbb P( \frac{S_n}{n} \to EX_1) = 1$$ and since $EX_1 \neq 0$, it means that $S_n \to \pm \infty$ almost surely. Meaning that starting some rank $N$, $\lvert S_n \rvert > x $ $\forall n \geq N$ almost surely $\implies \Bbb P(V = \infty) \leq \Bbb P(V>N)=0$.
Is this right ? I feel the reasoning is good but the proof can be re written $100$ times better (or other proof). Also does it apply to random walks in $\Bbb Z^d$ too ?