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.
Something may be wrong:
In fact, Assuming $X$ is $T_1$, if $x$ is a limit of $E$, for any open set $U$ of $x$ contain infinite number of points from $E$. So $E$ don't need to be compact.
Example: Let $X=[0,\omega_1)$. It is a countably compact space, since every limit point is in $X$. However it is not compact. Note that for every limit point $x$ of $X$, its every open nbhd contains infinite pints from $X$.
Here is something related your question which you may be interested in:
Let $X$ be a space and $A$ a subset of $X$. A point $x\in X$ is a point of complete accumulation of $A$ if $|U\cap A|=|A|$ for every open nbhd $U$ of $x$.
A space $X$ is compact iff every infinite set in $X$ has a point of complete accumulation.
Proof: see Here
Best Answer
Let $a$ be a point in $A$. Since $A$ is open, there is a number $\delta > 0$ such that $(a - \delta, a + \delta) \subset A$.
We now show $a$ is a limit point of $A$. To that end, let $\epsilon > 0$ be given. We may assume, without loss of generality, that $\epsilon < \delta$. Now, $(a - \epsilon, a + \epsilon) \subset (a - \delta, a + \delta) \subset A$, so $(a - \epsilon, a + \epsilon) \cap A = (a - \epsilon, a + \epsilon)$. Since $(a - \epsilon, a + \epsilon)$ intersects $A$ in a point other than $a$, we conclude $a$ is a limit point.