Understanding why if $\alpha = \sup E$ exists then $\alpha$ may or may not be an element of $E$ I'm reading Rudin Principles of Mathematical Analysis and am having trouble understanding this point. We haven't developed the real numbers yet (on the verge of doing so as the informed reader of this question is probably aware).

if $\alpha = \sup E \in Q$ exists then $\alpha$ may or may not be an element of $E$
Take $E_1$ to be the set of all $r \in Q$ with $r< a$ and let $E_2$ be the set of all $r \in Q$ with $r\le a$
$$\sup E_1 = \sup E_2 = a$$

I need external confirmation that my understanding of this statement is sound.
Specifically, I must show that $\sup E_1 = \sup E_2 = a$ by the definition of supremum given:

Definition
Suppose $S$ is an ordered set, $E \subset$ and $E$ is bounded above.
Suppose there exists an $\alpha \in S$ with the following properties
(I) $\alpha$ is an upper bound of $E$
(II) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E$
Then $\alpha$ is called the least upper bound of $E$ and we write
$$\alpha = \sup E$$
(Rudin, Walter. Principles of mathematical analysis. Vol. 3. New York: McGraw-Hill, 1964. 4. Print)

First, I'll verify that $a$ is an upper bound of $E_1$ and $E_2$. If $a \in Q$ is an upper bound then for all $b \in E_1, E_2$, $b \le a$. Inspecting how $E_1$ and $E_2$ are constructed verifies this.
What I do not understand is why the second part of the definition of supremum holds for $E_1$ and $E_2$. Please tell me if my reasoning is correct.
Here's my attempt:
For $E_1$:

If $\gamma < a$, we can choose $\gamma^\prime = \gamma + \frac{a-\gamma}{2} <a \implies \gamma^\prime \in E_1$ and $ a > \gamma^\prime > \gamma$. Therefore $\gamma < a$ is not an upper bound of $E_1$. This completes the lemma that $\sup E_1 = a$
Note 1 The reason the $E_1$ case is more complicated is because it is not clear that $\gamma < a \implies \exists x\in E_1 < \gamma$.

For $E_2$:

The verification is simple to see from taking $\gamma < a$.

Secondly, I'm hoping there is a better way of writing the verification than by my awkward construction of $\gamma^\prime$. In general, are moves like this bad or necessary. I just cannot see any other way to do it, re note 1.
 A: $*$Updated$*$
I checked Rudin's text and answered my question. I see there is an example involving zero.
So we consider the case $a=0$. Assuming you still want to proceed by contradiction, we shall assume $\hat{\gamma}$ is an upper bound for $E_1$ but $\hat{\gamma} < 0$. By the Archimedean Principle, we may choose $N \in \mathbb{N}$ so that 
\begin{equation}\displaystyle N>-\frac{1}{\hat{\gamma}} \;.\end{equation}
Since $-\frac{1}{N}$ is an element of $E_1$ we have reached a contradiction. Thus $0$ satisfies $(ii)$ from the definition of $\sup E_1$.
$*$for future reference$*$
Notice $(i)$ and $(ii)$ are equivalent to saying, for each $\varepsilon>0$ there exists $x \in E$ such that $\,\; \alpha - \varepsilon < x \leq \alpha$.
$\forall \varepsilon>0 \exists x \in E : \big\{ \alpha - \varepsilon < x \leq \alpha \big\}$
Knowing $\mathbb{Q}$ is dense in $\mathbb{R}$ is probably helpful in any problem dealing with a set such as $E_1$. Because it is helpful for getting your hands on this positive number $\varepsilon$ from this more concise definition. 
A: This is the (roughly) correct approach.

The average of two unequal numbers always lies strictly between them, and the average of two rational numbers is always rational.

These two assertions together tell you that there is some rational number ($\gamma'$) lying between $\gamma$ and $a$ which falls in the interval $(-\infty, a)$, if $a$ is a rational number. If you are not given that $a$ is rational, you need to find a rational number between the two, which can be easily justified using the density of $Q$ in $R$ or some moderately complicated algebra with floor functions.
This is not an awkward construction at all. Quite the opposite, taking the average of two numbers is the standard way  to produce a number between them.
