Let $X_1,X_2,\ldots$ be an iid sequence such that $P\{X_1 = 1\} = u$, $P\{X_1 = -1\} = d$ and $P\{X_1 = 0\} = 1-(u+d)$. We have that $E[X_1] \neq 0$.
Define $S_n = \sum_{i=1}^nX_i$ and $S_0 = 0$ and consider the natural filtration. Define the stopping time $\tau = \inf\{n\geq 0: S_n = a \text{ or } S_n = b\}$ where $a < 0 < b$ and $a,b \in \mathbb{Z}$. I need to show that $\tau$ is almost surely finite and the exercise asks me to use the law of large numbers (this is the part I have a problem with).
The LLN gives
$$\frac{1}{n}S_n \to u-d \quad \text{ on } \quad \Omega'$$ and $P\{\Omega'\} = 1$. I fix $\omega \in \Omega'$ and some $\varepsilon > 0$ (I feel like I need to make a clever choice for this later). Then, there exists $M_{\omega,\varepsilon}$ such that
$$n(u-d - \varepsilon) < S_n(\omega) < n(u-d+\varepsilon)$$ for every $n \geq M_{\omega,\varepsilon}$.
Now, if $u-d > 0$, then I pick $\varepsilon < u - d$ and I take $n$ large enough so that $n(u-d+\varepsilon) > b$. So then I know $S$ will hit $b$ on $\Omega'$. If $u-d < 0$, I pick $\varepsilon > u - d$ and so on.
Is this proof OK?