Open sets definition I find it really confusing to understand the open set definition which goes: 

$N$ is an open set in $E$ if for each $x\in N$, there exists $\varepsilon>0$ such that the open ball $B(x, \varepsilon)$ is contained in $N$

Could someone please try to break it down for me or maybe a graph.
It would all help 
 A: I will go a bit into the intuition in 2D. In the end I will explain how this connects back to 1D, because I assume $E$ is the real number line.

See the two pictures below. It is common to picture (at least in visual applications) the open sets with a dashed boundary, to indicate that these points (the boundary) does not belong to the set itself. So the left set should be closed, the right one should be open.
$\quad\quad\quad\quad\quad\quad\quad$
Because on the left, the boundary belongs to the set, you can choose $x$ on this boundary. Now ask yourself: is there a little circle (ball) around $x$ that is completely inside the set $N$? No of course not. At the boundary, moving only a little bit in the wrong direction will bring you outside.
In contrast, for the right image, the boundary does not belong to $N$. Thus it is not possible for you to choose $x$ on the boundary. But when $x$ is not on the boundary, it can wiggle a bit inside of $N$ without leaving it. Maybe just a tiny tiny bit. But this bit is sufficient to choose a circle $B(x,\varepsilon)$ around $x$ that is completeley contained in $N$. And this is what is so special about open sets. In any point you are sufficiently inside the set, so that you can wiggle without leaving.
On the left side, there are still point where you can wiggle without falling out of $N$, but the important thing for open sets is that this is true everywhere. And on the left side we found the boundary where this is not true.

In the 1D-case of $E$, you cannot imagine this balls as circles or balls, but you have to choose the appropriate 1D-analogon: intervals. Also, going with the anagoly of "wiggling a bit", you now can only wiggle in two directions: forth and back, and not up and down or whatever.
Sitting in $N=[0,1]$ at the point $x=1$, only moving a tiny tiny bit in direction of $2$ will bring you outside $N$. So $1$ is a boundary point of $N$ and $N$ is not open.
But when you are in $N=(0,1)$, you cannot sit in $1$ (or $0$), because these points do not belong to $N$. Instead, you will be at some point close to $1$ (or to $0$), say $1-\varepsilon$. At this points, you can wiggle in all direction by a bit of $\frac12\varepsilon$ without leaving $(0,1)$. Hence, the 1D-analogon of the ball $$B\left(x,\frac\varepsilon 2\right)=\left(x-\frac\varepsilon 2,x+\frac\varepsilon 2\right)$$ is completely inside $(0,1)$. This is true for all $x$ you can choose. Hence $(0,1)$ is open.
A: It means that around every point $x\in N$ you can find (read: draw if it makes it easier to understand, but this definition can be used in settings where drawing is not easy) a ball of positive radius $\varepsilon$ such that all points within the ball (that is all points near $x$) are within $N$. I.e. no point in $N$ is on the boundary. The whole point about $E$ is just that when considering openness in $E$ you should only consider points in $E$, 
A: Glad that you asked. Pick any $x \in (0,1)$. $0>x>1$. Take $\epsilon = \min\{x, 1-x\}$. You can check that the open ball $B(x,\epsilon)$ is contained in $(0,1)$.
