Show every chain has an upperbound? Sometimes I feel like proofs like this are pointless. I mean, if we have a partially ordered subset, it seems automatically true that you have a max element. 
1) Either you have an infinite sequence that converges to a point. Ie, all number < $\sqrt{2}$ and that element has an upperbound, or you have a finite collection of points and than it's obvious it has an upperbound. 
What is there to prove? 
 A: The real numbers are a partially ordered set. In fact a chain.
Is there an upper bound to the real numbers?
If you are asking in the context of applying Zorn's lemma to partial orders, there are two points here:

*

*We often apply Zorn's lemma to partial orders which involve sets that have properties "of finite character", namely if a set fails to have a property it is witnessed by a finite subset. Things like linear independence, etc.
So we pick which proofs we use Zorn's lemma in a manner that the proof fits the description of "easy application of Zorn's lemma".


*Often you apply Zorn's lemma after you had to show with some work that indeed every chain will have an upper bound independently. Then the proof itself is rather easy already.
A: Here are some chains in posets with no upper bound:


*

*The finite subsets of $\mathbb{N}$, ordered by inclusion:
$$\{ 0 \} \subseteq \{ 0, 1 \} \subseteq \{ 0, 1, 2 \} \subseteq \{ 0, 1, 2, 3 \} \subseteq \cdots$$

*The real numbers (or rational numbers or integers or natural numbers), with their natural ordering:
$$ 0 < 1 < 2 < 3 < \cdots$$

*The positive natural numbers $\mathbb{N}^+$, ordered by divisibility:
$$2 \mid 4 \mid 8 \mid 16 \mid 32 \mid \cdots$$

*The set of continuous functions $[0,1] \to \mathbb{R}$, ordered by declaring $f \le g$ if and only if $f(x) \le g(x)$ for all $x \in [0,1]$: define
$$f_n(x) = \begin{cases} nt & \text{if } 0 \le t \le \frac{1}{n} \\ 1\ & \text{if } \frac{1}{n} \le t \le 1 \end{cases}$$
then $f_1 \le f_2 \le \cdots$, but any least upper bound $g$ must have $g(0)=0$ and $g(t)=1$ for all $t>0$, so is discontinuous.
...and so on.
