$F \subseteq X$, where $X$ is a metric space.
Then $F$ is closed if and only if every convergent sequence in F converges to a limit in $F$.
Attempt
$ \implies$ If $F$ is closed, $F= \overline F$, but $\overline F=F \cup F'$ where $F'$ is the set of all accumulation points of $F$.
Thus, $F' \subseteq F$ which means that $F$ contains all of its limit points which means that every sequence in F converges to a limit that is in F.
Is this a valid proof for this direction?
Edit
After reading the comments, I should have written every convergent sequence in F converges to a limit in F.
Best Answer
That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).
For the converse: If every convergent sequence in $A$ converges to a point in $A$, then $A'\subset A$. Thus A is closed.