EDIT: That whole answer assumes the $p_n$'s are uniformly and independently distributed on $[0,1]$. I have no idea on how to extend it to the general case of other distributions.
It gets elementary after you look at $A$ in simple cases. For example when $\ell(x) = 0$ and $u(x) = \frac{1}{2}$ for all $x$ and the $p_n$'s are uniformly and independently distributed on $[0,1]$. Then almost surely both $p_n\in[0,\frac{1}{2}]$ for infinitely many $n$'s and $p_n\in[\frac{3}{4},1]$ for infinitely many other $n$'s. Therefore $d\big(p_n, [0, \frac{1}{2}]\big)$ keeps oscillating too widely and $$a := \lim_{n \to \infty}d\big(p_n, [\ell(x), u(x) ]\big)$$ almost surely doesn't exist and $A$ is almost always empty in that special case of functions.
Now let's go back to the general case of any real functions $\ell()$ and $u()$. With the same reasonning, $a$ almost surely doesn't exist for any $x$ with $0 < \ell(x)$ or $u(x) < 1$. Conversely, $a$ is always $0$ for any $x$ with $\ell(x) \le 0$ and $1 \le u(x)$. Therefore $$ A = \{ x\in X: \ell(x) \le 0 < 1 \le u(x)\} $$ and trivially $$A\subseteq A_n\quad \forall n.$$
EDIT: This proves (2) is true iff $\ell(x) \le 0 < 1 \le u(x)$ for some $x\in X$.
(1) and (3) are false when for example $X=[0,1]$ and $\ell(x) = 0$ on $[0,\frac{1}{2}]$, $\ell(x)=\frac{3}{4}$ on $(\frac{1}{2}, 1]$ and $u(x) = 1$ on $[0,1]$. Because then $A = [0,\frac{1}{2}]$ and $A_n$ almost surely contains $\frac{3}{4}$ for infinitely many $n$'s hence $d_H(A,A_n)>\frac{1}{4}$ infinitely often and cannot go to zero.
Along these lines, you should have no trouble showing (1) and (3) hold if the functions $\ell()$ and $u()$ are continuous on $X$, $\ell(x) \le 0 < 1 \le u(x)$ for some $x\in X$ and $\ell(x) > 0$ for some other $x\in X$.