$A_n=[n,\infty)$ in $\mathbb{R}$ with a Euclidean metric.
A set is closed if it contains all its limit points. A limit point is a point whose neighborhood contains a point in the set. I'm not sure what happens when we consider a neighborhood at infinity.
Is $A_n$ the set of limit points of $A_n$? If this is true, $A_n$ is closed.
Your definition of "limit point" is wrong. The expression "a point whose neighbourbood" makes no sense, since a point generally has more than one neighbourhood. Also, the point contained in the neighbourhood may not be the point itself. Rather, a limit point of a set is a point $x$ each of whose neighbourhoods contains a point in the set other than $x$.
It's also not entirely clear what you mean by "a neighbourhood at infinity". You may be confused by the notation $[n,\infty)$. Here the symbol $\infty$ is a notational device and not an element of $\mathbb R$. Points in $\mathbb R$ have "neighbourhoods at infinity" in the sense that that they have neighbourhoods that extend to infinity, but this is irrelevant here since whether a point is a limit point is decided "at close range": Since every superset of a neighbourhood contains all points of the neighbourhood, the condition that every neighbourhood must contain a point means that every arbitrarily small neighbourhood must contain a point.
In the present case, almost all points $x\in A_n$ are in the interior of $A_n$ and have entire neighbourhoods within $A_n$ (for instance $(\frac{x+n}2,\infty)$), except for $x=n$, but here, too, every neighbourhood contains an open set $(n-\epsilon,n+\epsilon)$, which contains the point $n+\frac\epsilon2\in A_n$.