I'm trying to prove Theorem 3.2.8 from Abbott's Understanding Analysis (2nd ed) which states that:
Theorem 3.2.8: A set $F \subseteq \mathbb{R}$ is closed if and only if every Cauchy sequence contained in $F$ has a limit that is also an element of $F$.
Here is my attempt at the proof which uses theorem 3.2.5 (A point $x$ is the limit point of a set $A$ if and only if $x = lim~a_n$ for some sequence $(a_n)$ contained in $A$ satisfying $a_n \neq x~ ,\forall n \in \mathbb{N}$)
Proof: ($\Rightarrow$) Suppose $F$ is closed and let $X$ be a set of all limit points and $X \subseteq F$. Suppose $x_k \in X$. To generate a Cauchy sequence $(a_n)$, consider $\epsilon = 1/n$ such that
$$
a_n \in V_{1/n}(x_k) \cap F \setminus \{x_k \}
$$
From theorem 3.2.5, $a_n \to x_k$ and since $x_k \in F$, $\Rightarrow$ is thus proved.
To prove ($\Leftarrow$), suppose ($a_n$) is a Cauchy sequence. Also suppose $x$ is the limit point of $F$. Since ($a_n$) is Cauchy, it is convergent and so from theorem 3.2.5, $x = lim~a_n$. Since, $x \in F$, this proves ($\Leftarrow$).
I would appreciate if someone can critique my proof writing and suggest if this proof is correct and rigourous