The basic idea is what could be called the monotonicity of $\sup$: the supremum over a set is at least as large as the supremum over a subset.
Of course, this only makes sense if the product of the $\limsup$s is not $0\cdot\infty$ or $\infty\cdot0$. We also make the assumption that $a_n,b_n\gt0$. To see that this is necessary, consider the sequences $a_n,b_n=(-1)^n-2$.
Recall the definition of $\limsup$:
$$
\limsup_{n\to\infty}a_n=\lim_{k\to\infty\vphantom{d^{d^a}}}\sup_{n>k}a_n\tag{1}
$$
The limit in $(1)$ exists since, by the monotonicity of $\sup$, $\sup\limits_{n>k}a_n$ is a decreasing sequence.
Furthermore, also by the monotonicity of $\sup$, if $a_n,b_n\gt0$,
$$
\sup_{n>k}a_n \sup_{n>k}b_n=\sup_{m,n>k}a_nb_m\ge\sup_{n>k}a_nb_n\tag{2}
$$
Taking the limit of $(2)$ as $k\to\infty$ yields
$$
\limsup_{n\to\infty}a_n\limsup_{n\to\infty}b_n\ge\limsup_{n\to\infty}a_nb_n\tag{3}
$$
since the limit of a product is the product of the limits.
If the limit of
$a_n$ exists, we have that for any
$\epsilon>0$, there is an
$N$, so that
$n>N$ implies
$$
a_n\ge\lim_{n\to\infty}a_n-\epsilon\tag{4}
$$
We are interested in small
$\epsilon$, so it doesn't hurt to assume
$\epsilon\lt\lim\limits_{n\to\infty}a_n$.
Thus, for $k>N$, if $a_n,b_n\gt0$,
$$
\sup_{n>k}a_nb_n\ge\left(\lim_{n\to\infty}a_n-\epsilon\right)\sup_{n>k}b_n\tag{5}
$$
taking the limit of $(5)$ as $k\to\infty$ yields
$$
\limsup_{n\to\infty}a_nb_n\ge\left(\lim_{n\to\infty}a_n-\epsilon\right)\limsup_{n\to\infty}b_n\tag{6}
$$
Since $\epsilon$ is arbitrarily small, $(6)$ becomes
$$
\limsup_{n\to\infty}a_nb_n\ge\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{7}
$$
Combining $(3)$ and $(7)$ yields
$$
\limsup_{n\to\infty}a_nb_n=\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{8}
$$
since $\displaystyle\limsup_{n\to\infty}a_n=\lim_{n\to\infty}a_n$.
You just go to the definition of $\limsup$. We define $A_n = \sup_{m\geq n} a_m$ and $B_n = \sup_{m\geq n}b_m$. Since $a_n \leq b_n$, $A_n \leq B_n$ for every $n>N$. Then, $\inf A_n \leq A_n \leq \inf B_n \leq B_n$, but $\inf A_n = \limsup a_n$ and $\inf B_n = \limsup b_n$.
Best Answer
Missing pieces:
You need to assume the sequences are bounded. Otherwise the statement may be even nonsensical: how to interpret $1\le 0\cdot \infty$, for example?
$(A_n)$ and $(B_n)$ are nonnegative nonincreasing sequences; these properties imply the existence of the limits $\lim A_n$ and $\lim B_n$, which allows to conclude about the existence and the value of $\lim (A_nB_n)$.
You could have assumed $a_n\ge 0$ instead of $a_n>0$.