Let $A$ be an ordered set and $A = \{(x,y) \in\mathbb{R}^2 | 1 \le x \le 2\}$ is bounded from above by $(3,0)$ in $\mathbb{R}^2$ (with lexicographical order) but $A$ has no supremum. Also, $A$ is bounded from below by $(0,0)$ but $A$ has no infimum.
Now, consider the example:
$U \supset A$ and $U = \{(-1,0),(0,0),(1,0),(2,0),(3,0),(3,1),(3,2)\}$.
I understand that $(0,0)$ is a lower bound of $A$ and also $(3,0)$ is an upper bound of $A$, but why $(0,0)$ is not the infimum and $(3,0)$ is not the supremum? For the sake of example, my arguments below are in contradiction with $A$ not having infimum nor supremum.
Let $I$ be the set of lower bounds of $A$.
$I = \{(-1,0),(0,0)\}$
By definition the infimum of $A$ is the greatest element in $U$ that is less than or equal to all elements of $A$.
$\inf(A) = (0,0)$
because $(0,0) \ge (-1,0)$ in the lexicographical order.
Similarly with the supremum:
Let $S$ be the set of upper bounds of $A$.
$S = \{(3,0),(3,1),(3,2)\}$
By definition the supremum of $A$ is the least element in $U$ that is greater than or equal to all elements $A$.
$\sup(A) = (3,0)$ because $(3,0) \le (3,1) \le (3,2)$ in the lexicographical order.
Best Answer
Let's look at the upper bounds of $A = \{(x,y) \in\mathbb{R}^2 | 1 \le x \le 2\}$. You already saw that $(3,0)$ is an upper bound, because $3>2$, but so are each $(3,-1)>(3,-2)>(3,-3)>\ldots$, so we need to find smaller upper bounds. The same holds if we replace $3$ above with some $a>2$.
Trying $a=2$ doesn't work, since for any $b\in\mathbb{R}$, we have that $(2,b)<(2,b+1)\in A$, so $(2,b)$ is not a canditate for the supremum. Obviously the there are no upper bounds with $a<2$.
So we are left with the set $\{(a,b)\mid a>2,b\in\mathbb{R}\}$ of upper bounds of $A$. The problem here is that we cannot find a smallest element in this set, as you can always choose $2+\frac{a-2}{2}$, which is smaller than $a$ (even if you could find this $a$, then we could still make $b$ approach $-\infty$, meaning that there is still no smallest element).
Analogously, $A$ doesn't have an infimum.