Your proof is mostly OK, except for the fact that it's not a proof of your original statement.
You are trying to prove that any set with greatest lower bounds automatically has least upper bounds. In your proof, you fix an arbitrary subset $B$ of a set $S$ with greatest lower bounds, and then show that a completely different set $A$ has a least upper bound. Instead, you need to show that $B$ has least upper bounds. Let's see how we can alter your proof so that it proves your original statement.
Let $B$ be a nonempty subset of $S$ that is bounded below.
This is (almost) a great place to start. If you're trying to prove something about all objects in a class, it is always good to start by fixing an arbitrary one. But here, you want to show that any non-empty bounded above subset of $S$ has a least upper bound, so you should be fixing $B$ to be a non-empty bounded above subset of $S$.
Let $a\in A$ be the set of all lower bounds of $B$.
First of all, I think you meant to say 'Let $A$ be the set of all lower bounds of $B$'. But even this is not what you want to do. Fixing $A$ to be the set of lower bounds of $B$ can't help you at all, since the only interesting thing you can say about $A$ is that it has a greatest element ($\inf B$), which is equivalent to $B$ having a greatest lower bound. You certainly can't apply the greatest lower bound property to $A$ - you don't even know that $A$ is bounded and $\inf A$, if it exists, is certainly unrelated to $B$.
At some point, you have to apply the greatest lower bound property to some set $C\subset S$, since you can't conclude that $S$ has the least upper bound property if you don't know that it has the greatest lower bound property (e.g., $\mathbb Q$ does not have the greatest lower bound property), but you have to apply it to a set that you know has a greatest lower bound; i.e., a set that is bounded below. Furthermore, it's going to have to be a set of elements that are $\ge$ the elements of $B$; otherwise, it'll have nothing to do with upper bounds of $B$ at all.
The solution, of course, is to fix $C$ to be the set of upper bounds for $B$. Now we're trying to show that $C$ has a least element.
There also exists an element $b\in B$ for all $a$ such that $a\le b$, so $A$ is bounded above (by $b$).
First of all, we aren't interested in saying that $A$ (or $C$) is bounded above, since being bounded above doesn't tell us anything. Being bounded below, of course, tells us that a set has a greatest lower bound. So we want to show that $C$ is bounded below. Secondly, it is ambiguous whether you mean that there is some $b$ that works for all $A$ or whether there is a separate $b(a)$ for each $a$. Both are true, but only the first implies your conclusion.
I'd prefer to write, 'Fix an arbitrary $b\in B$. Then for each $c\in C$, $c$ is an upper bound for $B$ so, in particular, $b\le c$. So $C$ is bounded below by $b$.'
Since $S$ satisfies the greatest lower bound property, $\inf \hspace{2 pt} B = \alpha$ exists for all $a$ such that $a \leq \alpha \leq b$.
This is where you started to go off the rails. Knowing that $B$ has a greatest lower bound can tell you nothing about whether it has a least upper bound. For example, the set $\{3,3.1,3.14,3.141,3.1415,3.14159,\dots\}$ has a greatest lower bound but no least upper bound in $\mathbb Q$.
I think I've given you enough clues for you to finish the proof on your own. For reference, here is my proof of the statement:
Let $B$ be a non-empty subset of $S$ that is bounded above, and let $C$ be the set of all upper bounds for $B$. Fix $b\in B$; then, for each $c\in C$, $c$ is an upper bound for $B$ so, in particular, $b\le c$. So $C$ is bounded below by $b$, which means that it has a greatest lower bound $\gamma$. Since every $b\in B$ is a lower bound for $C$, $\gamma\ge b$ for each $b\in B$. So $\gamma$ is an upper bound for $B$, and is clearly the least upper bound since it is a lower bound for the set of upper bounds. $\Box$
The proof is fine. You just need to realise that everything "lives" in $S$.
So $(S,<)$ is linearly ordered and satisfies the lub property. This means that every $B \subseteq S$ that is bounded above (which means: $\exists b \in S: \forall x \in B: x \le b$) then $B$ has a least upper bound. Now he wants to prove that $S$ has the glb-property. So for every $B \subseteq S$, if $B$ has a lower bound (so $\exists b \in S: \forall x \in B: b \le x$) there exists a greatest lower bound for $B$.
So if we have such a $B$ that is non-empty and bounded below by definition of being bounded below the set $L = \{b \in S: \forall x \in B: b \le x \}$ is non-empty. This is what being bounded below means in the ordered set $S$. And as $B$ is non-empty, pick $x \in B$. Then for every $b \in L$, by definition of being in $L$: $b \le x$. So $x$ (which is in $B \subseteq S$) shows that $L$ is bounded above (in $S$), and the rest of the proof goes through.
In your example, $S = (0, 2]$ and $B = (0,1]$ (for definiteness) in their usual order, the $S$ satisfies the lub-property, but the $B$ is not bounded below in $S$ (For every $x \in S$ , with $x < 2$, $\frac{x}{2} < x$ and lies in $B$. So $x$ is not a lower bound for $B$.). So we don't have to show that $B$ has a greatest lower bound, as it has no lower bound at all. So the example is irrelevant. It's not a counterexample to $S$ also having the greatest lower bound property.
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