Let \(X\) be a first countable space and \(A\subseteq X\). Then \(x\in \Int A\) if and only if every sequence converging to \(x\) is eventually in \(A\), i.e. at most finitely many points of the sequence are not in \(A\).
Proof
Assume \(x\in \Int A\) and \(x_n \to x\). Then due to the definition of convergence \((x_n)\) is eventually in \(A\).
Suppose on the contrary that \(x\not\in \Int A\). Then it is in \(\partial A\) or \(\Ext A\). Since \(X\) is fist countable there is a nested neigbourhood basis of \(x\). Using the characterization of the boundary and the characterization for open sets (and therefore \(\Ext A\)) we may construct a sequence by choosing elements in \(X\setminus A\) which converge to \(x\).