We denote the asymptotic cone of a set $S$ by $A(S)$: given $S\subset \mathbb{R}^n$ and a natural number $k$, let $\Gamma(S^k)$ be the smallest closed cone at $0$ containing $S^k=\{x\in S: ||x||\ge k\}$. The asymptotic cone of $S$ is $A(S)=\bigcap_{k\ge 0} \Gamma(S^k)$. Intuitively, $A(S)$ contains (the closure of) all directions $S$ contains at infinity.
For two non-empty sets $S\subset \mathbb{R}^n, T\subset \mathbb{R}^m$, it is claimed that $A(S\times T)\subset A(S)\times A(T)$.
Question 1: How to prove it?
Question 2: Is there an example such that $A(S\times T)\neq A(S)\times A(T)$?
Lemma: $x \in A(S)$ if and only if there exists a sequence $s_k \in S$ and $\lambda_k \in [0, \infty)$, converging to $0$, such that $x = \lim_{k \to \infty} \lambda_k s_k$.
Proof. Suppose $x \in A(S)$. Then $x \in \Gamma(S^k)$ for all $k \ge 0$. Choose $x_k$ in $\Bbb{R}^+S_k$, the (not necessarily closed, not necessarily convex) conical hull of $S_k$, such that $\|x - x_k\| \le 1/k$. We know that $x_k$ must take the form $\lambda_k s_k$ for some $\lambda_k \ge 0$ and $s_k \in S^k$.
If $\operatorname*{limsup}_k\lambda_k > 0$, then a subsequence $\lambda_{k_l}$ converges to some $\lambda > 0$. We then have, $$\|x_{k_l}\| = \lambda_{k_l} \|s_{k_l}\| \ge \frac{\lambda}{2}\|s_{k_l}\| \ge \frac{\lambda}{2}k_l \ge \frac{\lambda l}{2},$$ for sufficiently large $l$, so $\|x_{k_l}\|$ diverges to $\infty$. This contradicts $x_k \to x$ as $k \to \infty$.
Thus, $\operatorname*{limsup}_k\lambda_k = 0$. The limit inferior must be at least $0$, hence $\lim_k \lambda = 0$.
Conversely, suppose $s_k \in S$ and $\lambda_k$ are as in the lemma's statement. If any subsequence $s_{k_l}$ of $s_k$ is bounded by $M \ge 0$, then $\|\lambda_{k_l} s_{k_l}\| \le \lambda_{k_l} M \to 0$. But, at the same time, $\|\lambda_{k_l} s_{k_l}\| \to \|x\|$, hence $x = 0 \in A(S)$.
Otherwise, $\|s_k\| \to \infty$ as $k \to \infty$. We can therefore take a subsequence $s_{k_m}$ such that $s_{k_m} \in S^m$, and so $\lambda_{k_m} s_{k_m} \in \Gamma(S^m)$, for all $m \in \Bbb{N}$. Then, $$d(x, \Gamma(S^m)) = d(x, \Gamma(S^m)) - d(\lambda_{k_m}s_{k_m}, \Gamma(S^m)) \le \|x - \lambda_{k_m}s_{k_m}\| \to 0$$ as $m \to \infty$. But, since the $\Gamma(S^m)$ sets nest descendingly, we know that $d(x, \Gamma(S^m))$ is a montone increasing function, converging to $0$, consisting of non-negative entries. Thus, the only possibility is that $d(x, \Gamma(S^m)) = 0$, i.e. $x$ belongs to the closure of $\Gamma(S^m)$ (which is $\Gamma(S^m)$). That is, $x \in A(S)$.
With that lemma, the question is straightforward. If $(v, w) \in A(S \times T)$, then there exists some $\lambda_k \ge 0$, converging to $0$, and a sequence $(s_k, t_k) \in S \times T$ such that $(v, w) = \lim_{k \to \infty} (\lambda_k s_k, \lambda_k t_k)$. This, of course, translates to $v = \lim_{k \to \infty} \lambda_k s_k$ and $w = \lim_{k \to \infty} \lambda_k t_k$, which implies, by the lemma, that $(v, w) \in A(S) \times A(T)$.
To see why $A(S) \times A(T)$ need not be a subset of $A(S \times T)$, consider the following subset of $\Bbb{R}$: $$S = T = \{2^n : n \in \Bbb{N}\}$$ Note that $A(S) = A(T) = [0, \infty)$, as the set consists of non-negative numbers and is unbounded above.
Suppose, for the sake of contradiction, that $(1, 3) \in A(S \times T)$. Then there exists some sequence $(2^{n_k}, 2^{m_k}) \in S \times T$ and $\lambda_k \to 0$ such that $(1, 3) = \lim_k \lambda_k(2^{n_k}, 2^{m_k})$. The quotient of these two limits reveals: $$\lim_{k \to \infty} 2^{m_k - n_k} = 3.$$ But, $2^{m_k - n_k}$ takes values from the set $\{\ldots, 1/2, 1, 2, 4, \ldots\}$, and every possible value of $2^{m_k - n_k}$ is at least distance $1$ from $3$, which is a contradiction.