= Solution
The <hitting time> $T$ is a <stopping time> for the <natural filtration>, because $\{T\leq n\}=\bigcup_{k\leq n}\{S_k=b\}$. The integer-valued increments are at most one. A path started at zero therefore cannot reach or exceed $b$ without first visiting $b$, so
$$
S_{n\wedge T}\leq b\quad\text{for every }n.
$$
Define $Y_n=b-S_{n\wedge T}$. It is nonnegative and integrable: $|S_{n\wedge T}|\leq\sum_{j\leq n}|X_j|$. Moreover,
$$
S_{(n+1)\wedge T}-S_{n\wedge T}=\mathbf1_{\{T>n\}}X_{n+1}.
$$
The <indicator function> is $\mathcal F_n$-measurable, and $X_{n+1}$ is independent of $\mathcal F_n$ with zero <expected value>. Taking <conditional expectation> therefore proves directly that $(Y_n)$ is a nonnegative <martingale>.
The general result used here is that a nonnegative discrete-time <martingale> has a finite almost-sure limit; this is the <martingale convergence theorem>, since $\mathbb E Y_n=\mathbb E Y_0=b$ bounds its <L1 norm>. On $\{T=\infty\}$ it follows that $S_n=b-Y_n$ converges finitely. A convergent sequence of integers is eventually constant, so $X_n=S_n-S_{n-1}$ would eventually be zero on this <event>.
However, the independent <events> $\{X_n=1\}$ all have the same strictly positive <probability>. The second <Borel-Cantelli lemma> says that they occur infinitely often <almost surely>. This contradicts eventual constancy on any positive-probability subset of $\{T=\infty\}$. Thus
$$
\boxed{\mathbb P(T<\infty)=1.}
$$
Notice that no lower bound on the negative increments was needed.
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