I understand in your comment above to Jonas' answer that you would like these things to be broken down into simpler terms.
Think about limit points visually. Suppose you have a point $p$ that is a limit point of a set $E$. What does this mean? In plain terms (sans quantifiers) this means no matter what ball you draw about $p$, that ball will always contain a point of $E$ different from $p.$
For example, look at Jonas' first example above. What you should do wherever you are now is draw the number line, the point $0$, and then points of the set that Jonas described above. Namely draw $1, 1/2, 1/3,$ etc (of course it would not be possible to draw all of them!!).
Now an open ball in the metric space $\mathbb{R}$ with the usual Euclidean metric is just an open interval of the form $(-a,a)$ where $a\in \mathbb{R}$. Now we claim that $0$ is a limit point. How?
Given me an open interval about $0$. For now let it be $(-0.5343, 0.5343)$, a random interval I plucked out of the air. The question now is does this interval contain a point $p$ of the set $\{\frac{1}{n}\}_{n=1}^{\infty}$ different from $0$? Well sure, because by the archimedean property of the reals given any $\epsilon > 0$, we can find $n \in N$ such that
$$0 < \frac{1}{n} < \epsilon.$$
In fact you should be able to see from this immediately that whether or not I picked the open interval $(-0.5343,0.5343)$, $(-\sqrt{2},\sqrt{2})$ or any open interval.
Now let us look at the set $\mathbb{Z}$ as a subset of the reals. What you do now is get a paper, draw the number line and draw some dots on there to represent the integers. Can you see why you are able to draw a ball around an integer that does not contain any other integer?
Having understood this, looks at the following definition below:
$\textbf{Definition:}$ Let $E \subset X$ a metric space. We say that $p$ is a limit point of $E$ if for all $\epsilon > 0$, $B_{\epsilon} (p)$ contains a point of $E$ different from $p$.
$\textbf{The negation:} $ A point $p$ is not a limit point of $E$ if there exists some $\epsilon > 0$ such that $B_{\epsilon} (p)$ contains no point of $E$ different from $p$.
From the negation above, can you see now why every point of $\mathbb{Z}$ satisfies the negation? You already know that you are able to draw a ball around an integer that does not contain any other integer.
Best Answer
This is wrong. All you know about $p$ is that it has points of $E$ arbitrarily close to it. Each $N_k(p)$ will contain a point of $E$, but it doesn't have to contain "many" of them or to be filled with them completely. Points of $N_k(p)$ other than $p$ may well not have a sequence of $E$-points converging on them. Perhaps the only presence of $E$ inside a small neighborhood around $p$ is just a sequence of points converging on $p$, but not on any other point there.
OK. General advice. Try to think geometrically, with pictures. Especially with metric spaces, you should always imagine little open balls of radius $\epsilon$ or whatever around some point. "Open set" means every point comes with a little ball around it that's entirely in the set, too. So you should always imagine that ball and put it to use. "Closed set" means every point NOT in it comes with a little ball around it that's also DISJOINT from the set. Or alternatively "closed set" means that every point arbitrarily close to us must be part of us.
Try to see if you understand why what's being claimed is true geometrically. $E'$ is the set of limit points, and we want it to be closed. Means that we ought to be able to prove: if $x \notin E'$, there's a little ball around $x$ that's also entirely not in $E'$, i.e. contains no limit points of $E$. How do we find the little ball? Clearly we need to understand exactly what it means to not be a limit point of $E$.
$x$ is a limit point of $E$ if any ball around $x$ contains a point from $E$ (apart from $x$ itself - I won't bother to repeat this condition every time).
Now negate:
$x$ is not a limit point if there's some ball around $x$ that does not contain a point from $E$.
OK, so now that we understand that, we know that any $x \notin E'$ already comes with a handy open ball $B_\epsilon(x) $ around it that is $E$-free. That's what "not being a limit point" means! But what we need is at least on the face of it a little more: we need a ball around $x$ that is $E'$-free, meaning every $y$ in that ball has its own ball around it that is $E$-free.
Can it be the same ball?
If you don't see this, draw an $x$, a small (open!) disc around it that is completely $E$-free (apart from $x$ itself, possibly), see if you can prove (by drawing) that everything in the disc is $\notin E'$.
After you figure out what the picture looks like, translate to formalism.