What should be considered the "right" solution depends very much on what facts about $\mathbb{N}\,$ that you are allowed to use.
It is not unlikely that you have been told, and can use, the following basic property of $\mathbb{N}$:
Every non-empty subset of $\mathbb{N}$ has a smallest element.
This is sometimes called the least number property, or, in more fancy language, the fact that under the natural ordering on $\mathbb{N}$, the set of natural numbers is well-ordered.
If you are allowed to use the least number property, let $X\,$ be your subset of $\mathbb{N}$. Let $S\,$ be the set of all upper bounds of $X$. You know that $S\,$ is non-empty, since you were told that $X\,$ is bounded above.
It follows that $S\,$ has a smallest element $m$. We show that $m\,$ is the least upper bound of $X$.
Suppose to the contrary that $m\,$ is not the least upper bound of $X$. Then there is an upper bound $b\,$ for $X\,$ which is $<m$. But then $b \in S$. Since $b<m$, this contradicts the fact that $m\,$ is the smallest element of $S$.
Or else possibly what you know about $\mathbb{N}$ is the Induction Principle.
The solution given above can be rewritten in terms of that, but it is a little less clean. I can do it if you indicate that the tool I used is not part of your official toolchest.
It’s not true that all closed sets are countable unions of disjoint closed intervals: the middle-thirds Cantor set is a counterexample. It is closed and nowhere dense in $\Bbb R$, so it contains no non-trivial closed interval. That is, the only closed intervals that it contains are degenerate ones of the form $[x,x]=\{x\}$. And since it is uncountable, it isn’t a countable union even of these degenerate closed intervals.
Best Answer
Yes, you're right: if $A$ contained two or more points, $a, b \in A$ with, say, $a < b$, then $glb(A) \le a < b \le lub(A)$.
Here's a countable, bounded subset of the reals that does not contain its lub or glb:
$$ A = \left\{ - \frac n {n+1} \mid n \in \mathbb{N}\right\} \cup \left\{ \frac n {n+1} \mid n \in \mathbb{N}\right\} $$ $A$ is $\{\dots, -\frac 3 4, -\frac 2 3, -\frac 1 2, 0, \frac 1 2, \frac 2 3, \frac 3 4, \dots \}$. Clearly, $glb(A) = -1$ and $lub(A) = 1$, but neither is a member of $A$.