A point $p$ is a limit point of the set $E$ if every neighbourhood of $p$ contains a point $q \not= p$ such that $q \in E$. So there are no limit points in $\mathbb{Z} \in \mathbb{R}$. Is that why it's closed?
A point $p$ is an interior point of the set $E$ if there is a neighbourhood $N$ of $p$ such that $N \subset E$.
So every point in $\mathbb{Z} \in \mathbb{R}$ is an interior point. So shouldn't $\mathbb{Z}$ be open as well?
By definition, a set $F$ is closed in a topological space $X$ if its complement $X\setminus F$ is open. Notice that with the usual topology on $\mathbb{R}$, $\mathbb{R}\setminus\mathbb{Z}=\bigcup_{n\in\mathbb{Z}}(n,n+1)$, which is the union of open intervals (open sets).
Another way to look at it is, as you pointed out, by showing that $\mathbb{Z}$ contains all its limit points in $\mathbb{R}$. Let $\mathbb{Z}'$ be the set af all limit point of $\mathbb{Z}$ in $\mathbb{R}$. As you already noticed, $\mathbb{Z}'=\emptyset\subset\mathbb{Z}$. The conclusion follows from the obvious fact that $\emptyset\subset A$ for any $A\subset\mathbb{R}$.
limit points are used to describe the boundary. The easiest answer is that $\mathbb Z$ is closed in $\mathbb R$ because $\mathbb R \backslash \mathbb Z$ is open.
Note that $\mathbb{Z}$ is a discrete subset of $\mathbb{R}$. Thus every converging sequence of integers is eventually constant, so the limit must be an integer. This shows that $\mathbb{Z}$ contains all of its limit points and is thus closed.
