Here's a counterexample, taking advantage of the option to have or not have a 0 in a poset. Take the index set to be $\{1,2\}$. Take $\mathfrak A_1$ to be an atomless Boolean algebra minus its 0 element. It is well-known and easy to see that this is separable; if $a\neq b$, then one of $a-b$ and $b-a$ in the Boolean algebra is non-0 and serves as the $x$ required in separability. Let $\mathfrak A_2$ be the poset consisting of three elements, namely a least element 0 and two larger, incomparable elements $p$ and $q$. It is easy to check that this is separable. I claim that the product of these two separable posets is not separable. Let 1 be the top element of $\mathfrak A_1$. Then $(1,p)$ and $(1,q)$ are distinct elements of the product, so separability would require the existence of a pair $x=(x_1,x_2)$ such that $x\curlyvee(1,p)$ but $x\not\curlyvee(1,q)$ or vice versa; by symmetry, I can ignore the "vice versa" possibility. Since $x\curlyvee(1,p)$, there must be $y=(y_1,y_2)$ such that $y_1\leq x_1$, $y_2\leq x_2$, and $y_2\leq p$ (the remaining conditions, that $y_1\leq 1$ and $y$ is not least in the product, are automatic, the first because 1 is the top of $\mathfrak A_1$ and the second because $\mathfrak A_1$ has no least element). Then $y'=(y_1,0)$ satisfies $y'\leq y\leq x$ and $y'\leq(x_1,0)\leq(1,q)$, and $y$ is not least in the product (again because $y_1$ is not least in $\mathfrak A_1$). This contradicts the fact that $x\not\curlyvee(1,q)$.

EDIT: In response to a comment from Porton, suggesting that I misinterpreted a quantifier, I admit that this is a possibility, but I think the result I claimed, that a product of separable posets with least elements is separable, holds under either interpretation. First, the interpretation I had in mind (because it seems the more reasonable concept) is that separability means that $a=b$ follows from $(\forall x)(x\curlyvee a\iff x\curlyvee b)$. As in the question, I'll write $n$ for the index set, $\mathfrak A_i$ for the factor with index $i$, and $\prod\mathfrak A$ for the product. I'll also write 0 for the least element in any poset. To prove separability, I'll assume $a,b\in\prod\mathfrak A$ with $a\neq b$, and I'll find an $x\in\mathfrak A$ such that either $x\curlyvee a$ and $x\not\curlyvee b$ or vice versa. Since $a\neq b$, there is an index $i$ with $a_i\neq b_i$; fix such an $i$. Since $\mathfrak A_i$ is separable, choose some $y\in\mathfrak A_i$ such that either $y\curlyvee a_i$ and $y\not\curlyvee b_i$ or vice versa; without loss of generality, assume the former, and fix a non-0 $z\leq a_i$ with $z\leq y$. Define $x\in\prod\mathfrak A$ by setting $x_i=z$ and $x_j=0$ for all $j\neq i$. Then $x\neq0$ (because $x_i=z\neq0$) and $x\leq a$, so $x\curlyvee a$. Yet $x\not\curlyvee b$, because any $w\in\prod\mathfrak A$ that is below both $x$ and $b$ would have $w_i$ below both $x_i=z$ and $b_i$, which is absurd as $z\leq y$ and $y\not\curlyvee b_i$.

The alternative reading of the definition of separability is that the colon after the quantifier makes the scope of the quantifier the whole implication rather than the antecedent. Under this interpretation, the implication is true for whatever value of $x$ one chooses. I choose the value 0 (since we're assuming the posets have least elements 0) and infer the implication $(0\curlyvee a\iff0\curlyvee b)\implies a=b$. But by definition of $\curlyvee$, we never have $0\curlyvee$ anything, so it is true that $(0\curlyvee a\iff0\curlyvee b)$ because both sides of the biconditional are false. So we infer from our implication that $a=b$. Thus, under this reading of the definition, the only way for a poset with a least element to be separable is to have only one element. And that's preserved by products.

twoseparable posets is separable, then I think this is enough. – David Roberts Jul 11 '12 at 23:33