Regard $A_n$ as a set of lattice points by
$$2^x 3^y \mapsto (x, y) \in \mathbb{N}_0^2,$$
where $\mathbb{N}_0$ is the set of nonnegative integers. For two elements $(a, b), (c, d) \in \mathbb{N}_0^2$, we denote $(a, b) \le (c, d)$ if $a \le c$ and $b \le d$. Also we denote $(a, b) < (c, d)$ if $(a, b) \le (c, d)$ and $(a, b) \ne (c, d)$. Then the problem can be interpreted as:

for any $A_1, A_2, \dots \subset \mathbb{N}_0^2$, there exist $i, j \in \mathbb{N}$ satisfying $\forall x \in A_i, \exists y \in A_j$ s.t. $y \le x$.

In this case we will say that the collection $A_1, A_2, \dots$ is "nice".

A reduced set $A$ is a subset of $\mathbb{N}_0^2$ satisfying $\notin x, y \in A$ s.t. $x < y$. Since a reduced set may have at most one point in $x = a$ (resp. $y = b$), it is a finite set. Moreover, if a reduced set $A$ has $(a, b) \in \mathbb{N}_0^2$ then $|A| \le a + b + 1$ holds. Obviously, we may assume that each $A_i$ is a reduced set.

Lemma 1. Given a collection $A_1, A_2, \dots$, suppose that there are no points $p$ in $\mathbb{N}_0^2$ which are contained in infinitely many $A_i$'s. Then the collection is nice.

Before proving the Main lemma, we introduce a result for a special case.

Corollary 2. Given a collection $A_1, A_2, \dots$, if there exists $N \in \mathbb{N}$ such that $|A_i| \le N$ for all $i$, then the collection is nice.

Proof. (mathematical induction) In case $N=1$, it is easy. Assume $N \ge 2$. If the collection satisfies the condition in Lemma 1 then it is done. If the collection does not satisfy the condition, i.e., $\exists p$ which is contained in infinitely many $A_i$'s. Set $\{B_i \mid i \in \mathbb{N}\}$ be the collection of $A_i$'s containing $p$. From the induction hypothesis, $\{B_i - \{p\} \mid i \in \mathbb{N}\}$ is nice, which induces easily that $\{B_i \mid i \in \mathbb{N}\}$ is nice. Thus the collection $A_1, A_2, \dots$ is also nice. $\square$

Let us go back to the original problem. If $A_1, A_2, \dots$ satisfies the condition in Lemma 1 then it is done. Otherwise $\exists p$ s.t. $p = (a, b)$ is contained in infinitely many $A_i$'s, then each cardinality of $A_i \ni p$ is bounded by $|A_i| \le a + b + 1$ (recall that $A_i$'s are reduced sets). Thus from the corollary 2, the collection of $A_i$'s containing $p$ is nice, and so does the original collection $A_1, A_2, \dots$, which completes the proof.

Now we only left the proof of Lemma 1.

Proof. (Lemma 1) For nonnegative integer $h$, define $V_h := \bigcup A_i \cap \{(h, y) \mid y \in \mathbb{N}_0\}$. If $V_h$ is a finite set then by the hypothesis of Lemma 1, only finitely many $A_i$'s contribute to $V_h$. So we may delete this finitely many $A_i$'s.

1. Suppose $V_h$ is finite for all $h \in \mathbb{N}_0$. Among the points in $\bigcup A_i$ take $(a_0, b_0)$, where $b_0$ is the smallest $y$ coordinate of points in $\bigcup A_i$ and $a_0$ is the smallest $x$ coordinate of points with $y$ coordinate are all $b_0$. By relabeling, set $(a_0, b_0) \in A_1$. From the collection $A_1, A_2, \dots$, remove all the $A_i$'s (except $A_1$) which contribute to $V_0, V_1, \dots, V_{a_0-1}$. Note that only the finite number of $A_i$'s are removed. Thus the collection of remaining elements (including $A_1$) is nice (by taking $(a_0, b_0) \in A_1$ as $x \in A_i$).

2. Suppose there exists $h \in \mathbb{N}_0$ such that $V_h$ is infinite. Among these $h$, let $t$ be the smallest element. Since each $A_i$ is finite, infinitely many $A_i$'s contribute to $V_t$. Set $\{C_i \mid i \in \mathbb{N}\}$ be the infinitely many collection for $V_t$. Define $H_k := \{(x, k) \mid (x, k) \in \bigcup C_i\}$. If all the $H_k$ are finite, then $\{C_i \mid i \in \mathbb{N}\}$ is nice. So assume there exists $k \in \mathbb{N}_0$ such that $H_k$ is infinite. Among these $k$, let $s$ be the smallest element. Since each $C_i$ is finite, infinitely many $C_i$'s contribute to $H_s$. Set $\{D_i \mid i \in \mathbb{N}\}$ be the infinitely many collection for $H_s$. Note that $D_i \cap V_t \neq \emptyset$, $D_i \cap H_s \neq \emptyset$. Thus if $(t, d) \in D_1$ and $(c, s) \in D_1$ then the number of set $D_i$'s which contribute to $[t, c) \times [s, d)$ is finite. So removing these $D_i$'s (except $D_1$) from the collection $\{D_i \mid i \in \mathbb{N}\}$ yields an infinite subcollection $\{D_1\} \cup \{E_i \mid i \in \mathbb{N}\}$ — of $A_1, A_2, \dots$. By taking $(t, d) \in D_1$ or $(c, s) \in D_1$ as $x \in A_i$, we can prove that $\{D_1\} \cup \{E_i \mid i \in \mathbb{N}\}$ is nice.

$\square$