General Point set Topology: The Closure of a set is Closed

Question

I already looked at about 3 problems on here (listed below), but they all seemed to use metric spaces. I was trying to understand this proof (from armstrong's basic topology), which just works in a general topological space. Basically, Armstrong shows for A contained in a topological space X, cl(A) is closed because X - cl(A) is open. He first shows if we take an element x in X - cl(A) , we can find an open neighborhood U around it that contains no elements of A (I think this is clear by definition of the complement). Then he says U can't contain any limit points since it's open. I think I understand everything expect that: Why would being a neighborhood of every point mean U couldn't contain any limit/accumulation points of A:

Armstrong's proof

These are articles I referenced, but I couldn't generalize it not in a metric space:

Closure of a set is closed proof

Prove the closure is closed and is contained in every closed set

[Proof Verification]: The closure of a set is closed.

Answer 1

If $U$ contains a limit point $x$ of $A$, then $U$, being a neighboorhood of $x$, must contain elements of $A$ (by definition of limit points), contrary to the choice of $U$.

Answer 2

One definition of the closure is as the intersection of all closed sets containing $A$. Or, the smallest closed set containing $A$. Hence closed.

As to the part you didn't understand: a limit point of $A$ can't have a neighborhood disjoint from $A$ (by definition of limit point).

Answer 3

To show that $Cl(A)$ is closed, we can show that it’s complement $X-Cl(A)$ is open. Take any point $x$ in $X-Cl(A)$. Because $x$ is not in the closure of $A$, it is in particular not a limit point of $A$.

Therefore, there is some open neighborhood $U\ni x$ such that $U\cap A=\emptyset$. Since $x$ was arbitrary, this proves that $X-Cl(A)$ can be written as a union of open sets and is therefore open.