Let $X$ be first countable space $(1^0)$:$ U \subset X$ is open $\iff$ whenever a sequence ${x_n}$ converges to x in U ,then $x_n$ eventually in U}.
I was able to prove the fist direction , suppose U is open and $x_n$ converges to $x$ ,$ x \in U$ U is open implies U is a nhood of x by the definition of converges
hence, since $x_n$ converges to $x$ , so there exist $n_ {\circ}$ such that $n \ge n_ {\circ} \Rightarrow x_n \in U $, so $x_n $is eventually in $U$.
But i have a problem in proving the other direction ,if any one can help me ?
note: $X$ is a first countable then at each $x \in X$ we can find a countable nhood base $\{U_n : n \in \mathscr N \}$ such that $U_1 \supset U_2 \supset U_3 \supset ....$