[proofplan] Both directions are proved by contradiction. For the forward direction, if a point of $A^c$ has no open ball contained in $A^c$, we construct a sequence in $A$ converging to that point, contradicting closedness. For the reverse direction, if a sequence in $A$ converges to a point outside $A$, that point lies in the open set $A^c$, contradicting eventual membership in $A^c$. [/proofplan] [step:Prove that $A$ closed implies $X \setminus A$ is open] Assume $A$ is closed (every convergent sequence in $A$ has its limit in $A$). Let $x \in X \setminus A$. Suppose for contradiction that no $r > 0$ satisfies $B(x, r) \subseteq X \setminus A$. Then for every $n \in \mathbb{N}$, the ball $B(x, 1/n)$ contains a point $x_n \in A$. The sequence $(x_n)$ lies in $A$ and satisfies $d(x_n, x) < 1/n$, so $x_n \to x$. Since $A$ is closed, $x \in A$. This contradicts $x \in X \setminus A$. Therefore there exists $r > 0$ with $B(x, r) \subseteq X \setminus A$, and since $x \in X \setminus A$ was arbitrary, $X \setminus A$ is open. [guided] We want to show that every point of $X \setminus A$ is an interior point, i.e., each has an open ball entirely contained in $X \setminus A$. We argue by contradiction. Fix $x \in X \setminus A$ and suppose no $r > 0$ satisfies $B(x, r) \subseteq X \setminus A$. In particular, for each $n \in \mathbb{N}$, the ball $B(x, 1/n)$ intersects $A$. Choose $x_n \in A \cap B(x, 1/n)$. The sequence $(x_n)$ is constructed so that $x_n \in A$ and $d(x_n, x) < 1/n$. The distance bound $d(x_n, x) < 1/n \to 0$ forces $x_n \to x$ by the squeeze theorem. Now we use the hypothesis: $A$ is closed, meaning it contains the limits of all its convergent sequences. Since $(x_n) \subset A$ and $x_n \to x$, closedness gives $x \in A$. But $x$ was chosen from $X \setminus A$, so $x \notin A$ --- a contradiction. Therefore the assumption was wrong: there exists $r > 0$ with $B(x, r) \subseteq X \setminus A$. Since $x \in X \setminus A$ was arbitrary, $X \setminus A$ is open. [/guided] [/step] [step:Prove that $X \setminus A$ open implies $A$ is closed] Assume $X \setminus A$ is open. Let $(x_n)$ be a sequence in $A$ with $x_n \to x$ in $X$. Suppose for contradiction that $x \in X \setminus A$. Since $X \setminus A$ is open, there exists $r > 0$ with $B(x, r) \subseteq X \setminus A$. Since $x_n \to x$, there exists $N$ with $d(x_N, x) < r$, giving $x_N \in B(x, r) \subseteq X \setminus A$. This contradicts $x_N \in A$. Therefore $x \in A$, and $A$ is closed. [/step]