= Solution
For each $N\geq1$, put $C_N=\bigcup_{n\geq N}A_n$. Closure of a <sigma-algebra> under <countable unions> and <countable intersections> gives
$$
A=\limsup_{n\to\infty}A_n=\bigcap_{N\geq1}C_N\in\mathcal F.
$$
The <events> $C_N$ decrease to $A$. <Continuity of probability> from above applies because $\mathbb P(C_1)\leq1<\infty$. Also, $A_n\subseteq C_N$ whenever $n\geq N$, so
$$
\mathbb P(C_N)\geq\sup_{n\geq N}\mathbb P(A_n).
$$
Letting $N\to\infty$ proves
$$
\boxed{\mathbb P(A)=\lim_{N\to\infty}\mathbb P(C_N)
\geq\lim_{N\to\infty}\sup_{n\geq N}\mathbb P(A_n)
=\limsup_{n\to\infty}\mathbb P(A_n).}
$$
This <reverse Fatou inequality for events> requires no <independence>.
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