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.
Let $A$ be a set. Interior points of $A$ are points $a\in A$ for which there exists some neighborhood $U$ of $a$ such that for all $x\in U$, $x\in A$. On the other hand, isolated points of $A$ are points $a\in A$ for which there exists some neighborhood $U$ of $a$ such that for all $x\in U$ other than $a$, $x\notin A$. In a sense, they are two opposite extremes. A point can easily be neither: we need only that for all neighborhoods $U$ of $a$, there is some point in $U$ other than $a$ which is in $A$, yet not all points in $U$ are in $A$.
A simple example is the point $0$ in the set $[0,1]$. This is not an interior point of $A$, as any neighborhood of $0$ contains negative numbers, but is not isolated either, as any neighborhood of $0$ contains numbers in $[0,1]$. A similar but slightly more interesting example is the set $\{0\}\cup \{1,1/2,1/3,\ldots\}$, in which $0$ is also neither an interior nor an isolated point.
As you suspected, all points in $\mathbb Q$ are neither interior nor isolated, as for any $a\in \mathbb Q$ we have rational and irrational points distinct from but arbitrarily close to $a$ thus any neighborhood of $a$ contains distinct points in $\mathbb Q$ and not in $\mathbb Q$.
Best Answer
Given $[0,3]\cup (3,5)$, we will prove that the interior contains $(0,5)$. Let $b\in (0,5)$. Three cases:
$b<3$. Then there is a neighborhood $(b-\epsilon, b+\epsilon)$ containing $b$ within $[0,3]$. We may find $\epsilon$ by taking $\min(b/2,(3-b)/2)$.
$b=3$. $(1,4)\subseteq [0,3]\cup(3,5)$ works.
$b>3$. Similar to case 1.
Now prove that no other points are in the interior.