Does $\mathbb R^n$ have the least upper bound property? I don't know if this is an axiom for $\mathbb R^n$ (I know this property is an axiom for $\mathbb R$), and hence the question. Please note, I am considering the obvious partial ordering on $\mathbb R^n$ i.e. if $x,y\in \mathbb R^n$ with $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ then $x\leq y$ iff for all $i$, $x_i\leq y_i$ and $x<y$ iff $x\leq y$ and for some $i$, $x_i<y_i$.
More concretely, I am interested in the following question: If $A\subset \mathbb R^n$ is a bounded set, I would like to talk about $\sup A$ and $\inf A$, with regards to the mentioned partial order. Do they exist? And can I say that I can find sequences $a_m\in A$ and $b_m\in A$ such that $a_m\to \sup A$ and $b_m\to \sup B$? Will there be any problem in the topology considered?
 A: Yes, this follows quite directly form the least upper bound property for $\mathbb R$.
However you're not guaranteed to have a sequence converging to the $\inf$ or $\sup$. You can for example take the set of $(t, -t)$ where $|t|\le 1$. The majorants of this set is $(x,y)$ such that $x\ge 1$ and $y\ge 1$. The least of these is $(1,1)$, but you can't make $(t_n, -t_n)\to (1,1)$ regardless of what sequence $t_n$ you choose.
Normally in general metric spaces one don't use the least upper bound property as much as one does in real analysis. The reason is partly the above failure, but also that it requires that the set is ordered which is normally not needed for what lub property is used for. Instead one uses the concept of completeness, the property that each Cauchy sequence is convergent.
A: Yes, a direct product of complete lattices is a complete lattice. You can compute sup and inf component-wise.
No, the set $A=\{(0,1),(1,0)\}$ is separated from $(0,0)$ and $(1,1)$.
A: Yes and no. Consider the set $A=\{(x,-x)\in \mathbb R^2 \mid -1\le x \le 1 \}$. It has a supremum but it is not the limit of anything from the set.
This problem is precisely why in analysis one rarely considers such a partial ordering on the product space (of arbitrary metric spaces). The topology associated with the ordering is only very loosely related to the product topology. A related issue is that of product of metric spaces, and the issue is resolved by using a norm. The (say, standard) norm on $\mathbb R^2$ Gives the correct topology, and has the property you are looking for.
