$(X, \tau) $ be a topological space.
• $A\subset X$ is said to be closed( or topologically closed) if $X\setminus A\in\tau$
• $A\subset X$ is sequentially closed of for every sequences $(x_n) \subset A$ such that $(x_n) \to x $ in $(X, \tau) $ implies $x\in A$
Topologically closed implies sequentially closed
Proof: Let $(x_n)\subset A$ such that $(x_n)\to x $ in $(X, \tau) $.
Suppose $x\notin A$.Then $X\setminus A$ is an open set containing $x$ and $(x_n)\to x$ implies $X\setminus A$ contains all but finitely many $x_n$'s which contradict that $(x_n) \subset A$.Hence $x\in A$
Sequentially closed may not imply topologically closed
Let $X$ be any uncountable set and $\tau_{cof}=\{U\subset X : U=\emptyset \text{ or} | X\setminus U|<\aleph_{0} \}$
Then $(X, \tau_{cof}) $ is a topological space where all proper closed subsets are countable.
Let us choose $A\subset X$ uncountable. Then $A$ is not topologically closed.
But $A$ is sequentially closed. As $(x_n) \subset A$ with $(x_n)\to x$ in $(X, \tau) $ Implies $x_n=x$ forall but finitely many $n\in\Bbb{N}$ ( a sequence in co-countable space is convergent iff eventually constant) .Hence $x\in A$.
If $(X, \tau)$ is first countable topological space then $A\subset X$ topologically closed iff sequentially closed.
My Question :$(X, \tau)$ be a topological space such that $\forall A\subset X$ sequentially closed implies $A\subset X$ topologically closed. Does this implies $(X, \tau) $ first countable?