Prove that a set X is closed if and only if it contains its accumulation points. Problem: Prove that a set $X$ is closed if and only if $X^a \subset X$.
Proof: Let $X$ be a closed set and suppose that $x \in X^a$. Then $\exists \{x_n\} \subset X$ with $x_n \neq x $ for all $n\in\mathbb{N}$ such that $\lim_{n\to \infty}x_n=x$. By definition, $x$ is a limit point of $X$, and $x\in X$. Hence $X^a \subset X$.
Now suppose that $X^a \subset X$ and let $x$ be a limit point of $X$. Then $\exists \{x_n\} \subset X$ such that $\lim_{n\to \infty}x_n=x.$ If $\{x_n\}$ is an eventually constant sequence, then $\exists N\in\mathbb{N}$ such that $x_n=x$ for all $n \geq N$, so $x\in X$. Thus suppose that $\{x_n\}$ is not eventually constant. It may still be the case that $x_n=x$ for infinitely many $n\in\mathbb{N}$. Construct a subsequence $\{x_{n_k}\}$ such that $x_{n_k}\neq x$ for all $k\in \mathbb{N}$. If it were not possible to construct such a sequence, then $x_n \neq x$ for only finitely many $n\in \mathbb{N},$ which would imply that $\{x_n\}$ is eventually constant. We have that $\{x_{n_k}\}\to x$ because subsequences converge to the same limit as their parent sequences, so by definition $x$ is then an accumulation point. Thus $x\in X$ and $X$ is closed.
I would appreciate any feedback on this proof, especially the second part. Thanks!

Disclaimer: I know that this question has been asked elsewhere, but the provided answers use concepts that have not yet been introduced in the book I am studying. Specifically, I don't know anything about the relationship between open and closed sets right now. As such, I would like to try proving it using only the concepts that I know right now.
Let $X^a$ denote the set of accumulation points of $X$. For this discussion, assume $X \in \mathbb{R}$.
Definition 1: A point $x$ is a limit point of $X$ if and only if there exists a sequence $\{x_n\} \subset X$ such that $\lim_{n\to \infty} x_n = x.$.
Definition 2: A set $X$ is closed if and only if it contains all of its limit points.
Definition 3: A point $x$ is an accumulation point of $X$ if and only if there exists $\{x_n\} \subset X$ with $x_n \neq x$ for all $n\in \mathbb{N}$ and $\lim_{n \to \infty}x_n = x$.
 A: Both parts of the proof are sound and rigorous. The only constructive feedback I can give you regards minor and more subjective aspects of your proof. 
One general selection you can make that will assist in both of the comments I have is to define the set of limit points. Suppose you call the set of limit points $X^{l}$. It follows from your definition of a closed set that a set $X$ is closed iff $X^{l} \subset X$. This definition can simplify and clarify your work.
My first comment regards the following line: "By definition, $x$ is a limit point of $X$, and $x\in X$. Hence $X^a \subset X$." This is entirely accurate, but in my opinion seems a bit unclear. Specifically, the fact that $x \in X$ stems from the fact that $X$ is closed--i.e., $X^{l} \subset X$. Using our new definition, you can simply state that "By definition, $x \in X^{l}$, giving $X^{a} \subset X^{l}$. Since $X$ is closed, $X^{l} \subset X$, thus $X^{a} \subset X$. 
Similarly, when you prove the reverse side of the implication, you can begin by letting some $x \in X^{l}$, and conclude with $x \in X^{a}$ implying that $X^{l} \subset X^{a}$. From your initial assumption that $X^{a} \subset X$, you then have $X^{l} \subset X$, which indicates that $X$ is closed. Once again, these are very minor suggestions, but I think that they can enhance the clarity of a few areas in your proof, and condense them. 
