Why is the set of integers in $\mathbb{R}$ closed?

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.