= Solution
Use the nonnegative <martingale> $(Y_n)$ and <filtration> $(\mathcal H_n)$ established above. Fix a finite $n$ and let
$$
H=\{\max_{1\leq k\leq n}Y_k\geq\lambda\},\qquad
T=\min\bigl(\{1\leq k\leq n:Y_k\geq\lambda\}\cup\{n\}\bigr).
$$
For $j<n$, $\{T\leq j\}$ is the <union> of the threshold-crossing <events> up to $j$, so $T$ is a <bounded stopping time>. The <optional stopping theorem> for a <martingale> at a <bounded stopping time> gives $\mathbb EY_T=\mathbb EY_0=1$. On $H$ one has $Y_T\geq\lambda$; on its <complement> one still has $Y_T\geq0$. Therefore
$$
\lambda\mathbb P(H)\leq\mathbb EY_T=1,
\qquad
\boxed{\mathbb P(H)\leq\lambda^{-1}.}
$$
Only a <bounded stopping time> has been used, so no extra <uniform integrability> assumption at an unbounded hitting time is implicit. This is also the <Doob maximal inequality for a nonnegative submartingale> specialized to a mean-one <martingale>.
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