= Solution
For each $n$,
$$
\{\tau\leq n\}=\bigcup_{k=0}^n\{S_k\in\{a,b\}\}\in\mathcal F_n,
$$
since $S_k$ is <adapted>. This verifies directly that $\tau$ is a <stopping time>.
Set $d=b-a$. On $\{\tau>md\}$, the <random walk> is at some integer strictly between $a$ and $b$. If its next $d$ steps are all $+1$, it must hit $b$ within that block. These future steps are <independent> of $\mathcal F_{md}$ and have <probability> $2^{-d}$ of being all $+1$. Hence
$$
\mathbb P(\tau>(m+1)d\mid\mathcal F_{md})
\leq(1-2^{-d})\mathbf1_{\{\tau>md\}}.
$$
The <law of total expectation> and <mathematical induction> give the <geometric tail bound from a uniform escape probability>:
$$
\mathbb P(\tau>md)\leq(1-2^{-d})^m\longrightarrow0.
$$
Therefore \b[$\mathbb P(\tau<\infty)=1$]. In fact this also proves $\mathbb E\tau\leq d\,2^d<\infty$ by grouping the <tail-sum formula> into blocks of length $d$. No assertion of <independence> between successive survival <events> is needed.
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