Let \(X\) be a topological space . A subset \(A\) is open if and only if for every point \(x\in A\) there is a neighbourhood contained in \(A\).
Proof
If \(A\) is open choose \(A\) is an open neighbourhood of every point in \(A\).
Otherwise let \(A_x\) be the neighbourhood contained in \(A\). Then \(A=\bigcup_{x\in A} A_x\) and \(A\) is open.