= Solution
Write $B_n=\{X_n>a_n\}$. If $B_n$ occurs, the running maximum at time $n$ exceeds $a_n$, so $B_n\subseteq A_n$ and $\limsup_nB_n\subseteq A$.
For the reverse inclusion, fix an outcome for which only finitely many $B_n$ occur. Choose $N$ such that $X_m\leq a_m$ for every $m\geq N$. The finite prefix has a finite maximum $K=\max_{1\leq m<N}X_m$ when $N>1$; there is no prefix to check when $N=1$. Since $a_n\to\infty$, for every sufficiently large $n$ that prefix satisfies $K\leq a_n$. For the remaining indices $N\leq m\leq n$, the nondecreasing thresholds give
$$
X_m\leq a_m\leq a_n.
$$
Thus $\max_{m\leq n}X_m\leq a_n$ for every sufficiently large $n$, so only finitely many $A_n$ occur. Taking the contrapositive proves the reverse inclusion, and hence the pointwise <event> equality
$$
\boxed{\limsup_n\{\max_{m\leq n}X_m>a_n\}
=\limsup_n\{X_n>a_n\}.}
$$
This <exceedance equivalence for increasing thresholds> is deterministic. Its proof uses both monotonicity of $a_n$ and divergence to $+\infty$; it does not require <independence> or identical <probability distributions>.
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