Definition A explicitly talks about subsets of real numbers, but in fact is perfectly applicable to any partially ordered set. By contrast, your Definition B does not explicitly say it is about real numbers, but in fact is only about subsets of the entire real numbers with the usual ordering.
In fact, we don't just talk about "least upper bound" when dealing with partially ordered sets, we talk about "least upper bound in $X$".
Your argument for equivalence assumes in fact that you are dealing with all of $\mathbb{R}$: to see how it can fail, consider the totally ordered subset of the real numbers given by $X=[0,1)\cup\{2\}$. Let $S=[0,1)$. Then $\{2\}$ is a least upper for $S$ in $X$, but your argument breaks down, since for $\epsilon=\frac{1}{2}$ we have that every $s\in S$ satisfies $s\leq 2-\frac{1}{2}$; yet $2$ is the least upper bound of the set $S$ in $X$, even though it does not satisfy Definition 2. The reason is that although, within the set of real numbers, $2-\epsilon$ "works" as an upper bound for $S$, this element is not in $X$; not being a member of the club of elements of $X$, it does not qualify to be a least upper bound in $X$, let alone the least upper bound.
So, definition A is valid in arbitrary partially ordered sets. Definition B is only equivalent to Definition A in some contexts, e.g., when dealing with the set of all real numbers in their usual ordering.
As far as what your lecturer said, it's hard to go by second-hand reports. It is possible that he said that in a partially ordered set $X$, a subset $S$ may have more than one minimal upper bound: an element $b\in X$ is a minimal upper bound for $S$ if and only if $s\leq b$ for all $s\in S$, and if $a$ is an upper bound for $S$ and $a\leq b$, then $a=b$.
As to your example, it does not really work because "$a-\epsilon$" has no inherent meaning; it can only mean something if you have an operation defined that allows you to subtract $\epsilon$ from $a$; since your $a$ and $b$ are not actually a real numbers (at least, not under the usual ordering) you cannot talk about $a-\epsilon$ directly; you have to explain what that means. And I would say that $a$ and $b$ are minimal upper bounds rather than "least upper bounds", since they both fail Definition A (which is the one that is applicable in generality).
I also want to note that your argument that Definition A and B are equivalent is badly phrased (possibly because you are talking about (1) and (2) and don't specify if you are talking about definition A or definition B). You should assume that $b$ satisfies (A) and prove that it satisfies (B) (you essentially do this, but you are assuming that your set is all of $\mathbb{R}$, or at least dense), and then assume that $b$ satisfies (B) and prove it satisfies (A).
If A has the least upper bound property then there is A0⊂A with A0 bounded above
That'd be true if $A$ didn't have the least upper bound property. You can always find a set that is bounded above or below. The question is whether every such set will always have a least upper bound.
Example: Let $A_0 = \{q| q^2 < 2\} \subset \mathbb Q$. That IS bounded above by $2$. But $2$ is not the least upper bound. There is no least upper bound for that set.
Also not having the least upper bound property doesn't mean that a least upper bound doesn't exist.
Example: Let $B = \{q| q < 3\}\subset \mathbb Q$. $\sup B$ does exist and $\sup B = 3$. Even though $\mathbb Q$ doesn't have the least upper bound property. In $\mathbb Q$, $B$ has a least upper bound.... and $A$ does not..
c=sup{A0}∈A, with x≤c for any x∈A0⊂A
Okay. $A$ does have the least upper bound property so $c = \sup A_0$ does exist. That's true.
Now consider the singleton subset A1={c}⊂A, consisting of only the element c, which is bounded below.
That'd be true of any singleton set, whether $A$ had the least upper bound principal or not.
Consider $A = \mathbb Q$ and $A_0=\{q| q < 3\}$ and $\sup A_0 = c$.
Now let $A_1 = \{c\} = \{3\}$.
Nothing interesting is going to come from this.....
Assuming that A does not have the greatest lower bound property, then inf{A1}∉A
Nonsense. $\mathbb Q$ does not have the greatest lower bound property but $\inf \{3\} = 3$.
Not having the greatest lower bound doesn't mean a greatest lower bound can't exist. It just means it doesn't have to exist.
$B=\{q| q^2 > 3\} \subset Q$ is such that $\inf B$ does not exist.
But $\inf A_1 = \inf \{c\} = c$ does.
Best Answer
Your final paragraph outlines a perfectly reasonable strategy for (a). So let's do that. And remember the defining property of $\sup$ (and analogously for $\inf$): It is the least upper bound, meaning all other upper bounds are larger. So any inequality of the form $\sup(X) \leq t$ is best proven by showing that $t$ is an upper bound for $X$.
The $\inf$ proof is entirely analogous, except we flip all the inequality signs, and change "upper" into "lower" (same goes for (b) and (c) as well).
Now for (b). This time there is only one inequality. It's rather simple to prove, though, using the defining property: We have that $\sup(B)$ is an upper bound for $B$, so it must be an upper bound for $A$.
The proof for (c) is very similar to half the proof for (a): again by using (b), clearly $\sup(A\cap B)\leq \sup(A)$, and just as clearly, $\sup(A\cap B)\leq \sup(B)$. Thus $\sup(A\cap B)$ is a lower bound for $\{\sup(A), \sup(B)\}$. (One could go the other way too, showing that $\inf\{\sup(A), \sup(B)\}$ is an upper bound for $A\cap B$. That would be more akin to the other half of the proof of (a). This time there isn't equality, because the two halves show the same inequality, rather than opposite inequalities the way they did in (a).)
Finally, we have (d). They give one hint as to what you could look at, but I like to do it simpler. Let us say our partially ordered set has three elements, two are incomparable, and the third is larger than both (the power set on a set of two elements yields a similar example). Then let $A$ and $B$ each consist of one of the two incomparable elements. Then the set $\{\sup(A), \sup(B)\}$ has no maximum as its elements are incomparable, by design.