Show that $\alpha\leq \beta$ 
Let $E$ be a non-empty subset of an ordered space. Suppose that a is a
  lower bound of $E$ and $\beta$ is a upper bound of $E$. Show that $
 \alpha \leq \beta $.

Proof:
(1) If $\alpha$ is a lower bound of $E$ then $\forall x\in E\quad x\geq \alpha$
(2) If $\beta$ is a upper bound of $E$ then $\forall x\in E\quad x\leq \beta$
From (1), (2) and since $E$ is ordered then
$$\alpha\leq x\leq\beta\Rightarrow \alpha\leq\beta$$ 
I'm starting now with real analysis and I'm still learning the art of demonstrating. Is this right? That's enough?
 A: It's almost enough. But no, I wouldn't give you full marks for it, because if you are just beginning, it's good to be strict.
And a strict judge is a stupid judge, meaning he sees what you wrote, not what you meant to write. So, a strict judge would look at your proof and ask a simple question:

What is this $x$ that popped up all of a sudden? For which $x$ is the inequality $\alpha\leq x \leq \beta$ true?

Another question I might ask here is (and this is already a hint on how to improve the proof):

Where did you use the fact that $E$ is non-empty? An empty set $E$ also satisfies all your conditions, yet the conclusion does not hold for it, so obviously there's something wrong with your proof!

A: The principle you are using is correct. If $E$ is an ordered set then "$\le$" has the transitivity property. Transitive means: $x \le y, y \le z \implies x \le z$ for all $x,y,z \in E$.
In terms of the proof you should always define your variables, so what is $x$?
Here is my example proof:
Since $E$ is non-empty, let $x \in E$. Since $\alpha $ is a lower bound for $E$, $\alpha \le x$. Since $\beta$ is an upper bound for $E$, $x \le \beta$. So by transitivity, $\alpha \le \beta$. 
