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.
The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property.
Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = \{x \in A: \forall b \in B: x \le b\}$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 \in B$ ($B$ is non-empty), and any $l \in L(B)$ we have $l \le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = \operatorname{lub}(L(B))$ exists. Claim: $l_0 = \operatorname{glb}(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b \in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 \le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m \in L(B)$.
As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m \le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = \operatorname{glb}(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $\operatorname{lub}(B) = \operatorname{glb}(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
Best Answer
Suppose that you’ve shown that every non-empty bounded subset of $\Bbb N$ has a maximum element. Let $A$ be a non-empty bounded set of integers, and let $n$ be a lower bound for $A$. Then $A+n=\{a+n:a\in A\}$ is a bounded subset of $\Bbb N$ and therefore has maximum element $m$; show that $m-n=\max A$.
Next, $-A=\{-a:a\in A\}$ is a bounded set of integers (why?), so it has a maximum element $m$, and it’s not hard to show that $-m=\min A$. Thus, it all boils down to showing that every non-empty bounded subset of $\Bbb N$ has a maximum element.
To show that every bounded subset of $\Bbb N$ has a maximum element, suppose that $A\subseteq\Bbb N$ is non-empty and bounded. Let
$$B=\{k\in\Bbb N:k\text{ is an upper bound for }A\}\,;$$
by hypothesis $B\ne\varnothing$, so $B$ has a least element $m$. Show that if $m\notin A$, then $m-1\in B$, contradicting the choice of $m$.