Q:

Topology : Neighborhood and limit points

I just started reading topology from "topology without tears" by SIDNEY A. MORRIS
Im stuck on the proof of "Let A be a subset of a topological space (X,
τ ).A point x ∈ X is a limit point of A if and only if every neighbourhood of x contains a point of A different from x".
While the proof of second statement assuming first is straightforward.I'm confused about the converse.
Definition of neighborhood is specified as: N is said to be neighborhood of x if there exists a open set U such that U ⊆ N and x belongs to U.
So my question here is if we assume converse to be true, then for every neighborhood of x, there exists an open set U such that U ⊆ N and N∩A is not empty set.But to prove x is a limit point of A, we need to show U∩A is not empty.
I'm unable to figure how.

A:

Assume that every neigbourhood of $x$ contains a point from $A$ different of $x.$ Since open sets containing $x$ are neigbourhoods of $x$ they contain a point of $A$ different from $x.$ This shows that $x$ is a limit point of $A.$

