For first countable topological space X we know if $x\in \bar{A} \iff \text{exist}\ x_n\to x$ with $x_n\in A$
Hence if $A$ is closed ,$x\in A $ there must exist a sequence converge to it.
My question is does the following statement holds?:
if for any $x\in A$ exist $x_n\to x$ with $x_n\in A$ then $A$ is closed?
It seems not,since closed iff contains all the limit point,there may exist some sequence $y_n\to y\notin A$. correct?
No. The sequence "for any $x \in A$, exists $x_n \rightarrow x$ with $x_n \in A$" is always true, since you can consider the constant sequence $x_n = x$.