I am working on this problem, and I am not 100% sure with my argument. Could someone help me verify it?

Let (X,d) be a metric space. Suppose that $\displaystyle <x_n>$ is a sequence in X such that it has no convergent subsequence. Claim: $\displaystyle A=\{x_1,x_2,x_3,...\}$ is closed in X.

By assumption, for every sequence $\displaystyle <p_n>$ in A, $\displaystyle <p_n>$ does not converge. Otherwise, $\displaystyle <x_n>$ has a convergent subsequence. This implies that A is closed since I'll show that a set A is closed in (X,d) iff for every sequence $\displaystyle <p_n>$ in A if $\displaystyle <p_n>$ converges then $\displaystyle lim p_n \in A$.

First direction: Suppose A is closed. Let $\displaystyle <p_n>$ be a convergent sequence in A. Assume $\displaystyle lim p_n=p \notin A$, then $\displaystyle lim p_n \in X-A$. But X-A is open and $\displaystyle lim p_n \in X-A$ implies that there exists N such that if n>N, then $\displaystyle p_n \in X-A$. Contradicts to our assumption that $\displaystyle <p_n>$ is a sequence in A.

Second direction: I'll show X-A is open. Let $\displaystyle <p_n>$ be a sequence in X converging to $\displaystyle p \in X-A$. It suffices to show that there exists N such that if n>N, then $\displaystyle p_n \in X-A$. Assume $\displaystyle p_n \in X-A$ for only finitely many times, then $\displaystyle <p_n>$ is a sequence in A. By hypothesis, $\displaystyle lim p_n \in A$. But $\displaystyle lim p_n=p \in X-A$. Contradiction. So, X-A is open, ie A is closed.