From your proof attempt, I see that you probably have the following definition of the closure of $E$ (which is a standard one):
$$\overline{E} = \{x \in \mathbb{R}: x \text{ adherent to } E\}$$
where $x$ is adherent to $E$ iff for all $\varepsilon>0$, there exists some $a \in E$ such that $|x - a| < \varepsilon$. (The last definition can be generalised to any metric space, or even any topological space as well, using open balls resp. open sets.)
It is clear that if $A \subseteq B$, then $\overline{A} \subseteq \overline{B}$, because the same point that would be used (for some $\varepsilon$) to show $x$ is adherent to $A$ also shows it's adherent to $B$.
As clearly $E \subseteq \overline{E}$, this implies $\overline{E} \subseteq \overline{\overline{E}}$
So suppose $x$ is adherent to $\overline{E}$. Let $\varepsilon > 0$. For the number $\frac{\varepsilon}{2} > 0$ we can apply that we know that $x$ is adherent to $\overline{E}$, so there is some $b \in \overline{E}$ such that $|x- b| < {\varepsilon \over 2}$.
Now as $b$ is adherent to $E$, in turn we find a point $a \in E$ such that $|b -a | < {\varepsilon \over 2}$. So in all $|x - a| \le |x - b| + |b -a | < {\varepsilon \over 2} + {\varepsilon \over 2} = \varepsilon$, which shows that (as $\varepsilon$ was arbitrary) that $x \in \overline{E}$.
So we also have $\overline{\overline{E}} \subseteq \overline{E}$.
The ${\varepsilon \over 2}$-idea was sound, but the sequence is unnecessary.
The last statement is easy: if $X \subseteq Y$ and $Y$ is closed, then by the first remarks above we get that $\overline{X} \subseteq \overline{Y} = Y$, the last equality because $Y$ is closed. So any closed set that contains $X$ also contains its closure.
