Conversely, suppose A ∩ cl(X A) = ∅. Let x be a point in A. Then x ∉ cl(X A), and hence there exists an open neighborhood U of x such that U ∩ (X A) = ∅. This implies that U ⊆ A, and hence A is open.
Here are some problem solutions from Engelking’s book on general topology: Let X be a topological space and let A be a subset of X. Show that the closure of A, denoted by cl(A), is the smallest closed set containing A. General Topology Problem Solution Engelking
General Topology Problem Solution Engelking** Conversely, suppose A ∩ cl(X A) = ∅
Let A be a subset of X. We need to show that cl(A) is the smallest closed set containing A. This implies that U ⊆ A, and hence A is open
Suppose A is open. Then A ∩ (X A) = ∅, and hence A ∩ cl(X A) = ∅.
Next, we show that A ⊆ cl(A). Let a be a point in A. Then every open neighborhood of a intersects A, and hence a ∈ cl(A).