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
Let $X$ be $[0,1] \cup \{2\}$ and $E$ be the intersection of $X$ with $\mathbb{Q}. $ Then $2$ is not a limit point but $E$ is dense.
It might be reasonable to define a limit point of $E$ to be $x$ such that there is a sequence $e_1,e_2,\cdots$ from $E$ with limit $x.$ That would make any point of $E$ a limit point of $E,$ the definition of a dense set could be briefer, and the answer to your question would be no. Equivalently: if we said every neighborhood of $x$ contains a point of $E$
However the definition is that the $e_i$ are distinct from $x$ (I.e. every neighborhood contains a point $e$ distinct from $x.$
It doesn't change what you can do, it just changed how you phrase things .
IMPROVED For the second question, an easy answer is $E=X=\mathbb{Z}.$ This is essentially the same as the discrete metric alluded to in another answer, $E=X$ and $d(x,y) \geq 1$ when $x \neq y.$
Here is an example $X=F\cup E$ living in $\mathbb{Q}^2$ consisting of $F=[0,1] \cap \mathbb{Q} $ together with $E,$ a countable cloud of isolated points whose set of limit points in $\mathbb{Q}^2$ is exactly $F.$
I'll comment at the end that we can easily modify the example to have an example living in $\mathbb{R}^2$ where $E$ is a different countable cloud of isolated rational points but the set of limit points is the entire real line.
Here is the description of a familiar type of construction for $E.$ The basic idea is to build $E$ as a sequence of points $(x_1,y_1),(x_2,y_2),\cdots $ in such a way that every point of $F$ is the limit of a sub-sequence from $E$ and each point $(x_i,y_i)$ is the center of a small disk of radius $r_i$ with no previous or future points of $X$ in it. If this seems familiar, skip to generalizations
Start with a specific enumeration $q_0,q_1,\cdots$ of the rational numbers in the unit interval. One, is to put $\frac{a}b$ before $\frac{c}d$ if either $a+b\lt c+d$ or $a+b=c+d$ but $a \lt c:$ $$\frac01,\frac10,\frac11,\frac12,\frac21,\frac13,\frac31,\frac14,\frac23,\frac32,\frac41,\frac15,\frac51,\cdots$$
Let $(x_i,y_i)=(q_s,2^{-i})$ where $i=2^s(2t+1).$ Then each $q_s$ is $x_i$ infinitely often and the corresponding $y_i$ converge to $0.$ Hence each point of $F$ is a limit point of a sub-sequence of $E.$ It is not hard to see that any limit point $(x,y) \in \mathbb{Q}^2$ of $E$ has $y=0$ ,so no point of $E$ is a limit point. Incidentally, No other point $(x,y) \in \mathbb{Q}^2$ can be a limit point of $E$ on account of the first coordinate.
GENERALIZATIONS Situating the example in $\mathbb{R}^2$ it is easy to adjust so that $F$ is the entire $x$-axis: First, with the existing $E,$ but situated in $\mathbb{R}^2,$ the set of limit points is all of $[0,1].$ (For $(x,0)$ with $0 \leq x \leq 1$ real just pick a rational sequence $q_{s_1},q_{s_2},\cdots$ conversing to $x$ and choose the points $(x_i,y_i)=(q_{s_j),2^{-i})$ for the $i=2^{s_j}.$
If we instead make $q_0,q_1,\cdots$ be an enumeration of all of $\mathbb{Q}$ and keep everything else the same, then the limit set of (the new) $E$ in $\mathbb{R}^2$ is the entire real line.
AN EVEN NICER EXAMPLE I should have remembered this. $E$ is a subset of $\mathbb{Q}^2$ which is dense in $\mathbb{R}^2$ and $X=E \cup Y$ where $Y=\mathbb{R}^2-Z$ with $Z$ a (the union of) a set of disks (one centered at each member of $E$) with total area less than $\epsilon.$ Let ${p}_i=(x_i,y_i)$ be an enumeration of all of $\mathbb{Q}^2.$ Consider the closed disk $D_i$ of radius $r_i=\frac{\epsilon}{2^i}$ around each ${p}_i.$ We may or may not use a particular ${p}_i$ for $E$ in which case we will or will not use the disk $D_i$ as part of $Z.$ To build $E$ and $Z$ start with both empty and consider the ${p}_i$ in order. If ${p}_i$ is not in the current $Z$ then update $E$ to include ${p}_i$ and enlarge $Z$ to $Z \cup D_i.$ Whatever $Z$ ends up being it has total area less than the combined area of all the $D_i$ which is less than $\epsilon.$ Each point of $E$ is the center of a closed disk otherwise devoid of any past or future points from $E$ and hence not a limit point.
Finally,it remains to show that for each point ${y} \in Y$ and $0 \lt \alpha $ there is a point of $E$ at distance no more than $\alpha$ from ${y}.$ There are only finitely many $D_i$ with $r_i \gt \frac{\alpha}2$ so exclude them and their disks $D_i.$ This is a finite union of closed disks so all rational points with $d(y,{p})$ small enough are outside that union. Pick a rational point with $d({y},{p}') \lt \frac{\alpha}2$ that small. Then there is a point ${p}_i \in E$ such that $D_i$ contains ${p}'$ Now $d({f},{p}') \lt \frac{\alpha}2$ ,and $d({p}',{p}_i)\lt \frac{\alpha}2$, so the point ${p}_i \in E$ is what was needed.