Proof about subset of $\mathbb{R}$ and closed interval. 
Let $E$ be subset of $\mathbb{R}$ and let $\{ I_j \}_{j=1}^m$ be a sequence of closed intervals such that $E \subset \cup_{j=1}^m I_j$.
Then, prove $\overline{E} \subset \cup_{j=1}^m I_j$. ($\overline{E}$ means closure of $E$.)

My attempt is following.
Pick up $x \in \overline{E}$.
If $x \in E$, then $x \in \cup_{j=1}^m I_j$ because $E \subset \cup_{j=1}^m I_j$ so $\overline{E} \subset \cup_{j=1}^m I_j$.
If $x \notin E$, then $x \in \overline{E} \setminus E=\partial E$.
I can't proceed. I don't use the fact that each $I_j$ is closed interval yet. But I don't know how I have to use the fact. I would like you to give me some ideas.
 A: First, it's not true that $\overline{E}\setminus E=\partial E$. Moreover, you didn't mention how you define $\overline{E}$. Anyway... $\overline E$ is the smallest closed set that contains $E$. Since $\bigcup_{i=1}^m I_i$ is closed and contains $E$, then $\overline{E}\subset \bigcup_{i=1}^m I_i$.
A: You'd like to use the fact that each $I_j$ is a closed interval. Since there are only finitely many $I_j$, their union is a closed set, although maybe not an interval. Call it $F$.
Next you wish to choose $x \in \overline{E} \setminus E \subset \partial E$, and show that $x \in F$ must be. In words, $x$ is a limit point for $E$, but one that does not belong to $E$.
Any neighborhood of $x$ intersects $E$. By this, you can have a sequence $z_k$ of distinct points of $E$, converging to $x$. We can simply view the $z_k$ as a sequence of distinct points of $F$, converging to $x.$
Each limit point for $F$ belongs to $F$, since $F$ is closed.
But our sequence $z_k$ shows that $x$ is a limit point for $F$, so we are done.
It is helpful to remember that for any set $A$ in any metric space, its closure has at most two types of elements. (Although it need not have any elements.)
These are isolated points of $A$ versus limit points for $A$. These two classes are mutually exclusive.
Summarizing our proof, note that the intervals were not important. We have shown

When $F$ is closed and $E \subset F$, then $\overline{E}\subset F.$

It is easy to write an example where $E \subset F$ is true but $\overline{E}\subset F$ is not.
Another proposition to consider is

$E \subset F\,\,\,$ implies(?) $\,\,\,\overline{E}\subset \overline{F}.$

