This same question was found here: Prove that F $ \in \mathbb{R} $ is closed if and only if every Cauchy sequences contained in F has a limit that is also an element of F. The OP / responses took a different path than what I am presenting here. I am presenting my own proof to see what could be worked on.
"A set F $\subseteq \mathbb{R}$ is closed $\textbf{iff}$ every cauchy sequence contained in F has a limit that is also an element of F."
$\textbf{Theorem 3.2.5}$: limit points of a set imply that the set has a convergent sequence to that limit point
$\textbf{Theorem 2.6.2}$: Cauchy sequences are convergent
(Both of the above theorems are "if and only if" theorems)
$\textbf{My attempt}$
($\underline{forward}$) Assume a set $F \subseteq \mathbb{R}$ is closed. Note that if F is empty we have a trivial case. So, lets assume F is non-empty and that there exists a Cauchy sequence, $(a_n)$, in F. By Theorem 2.6.2, $(a_n)$ converges to some limit, call it a. By definition, "a" is a limit point. Since F is closed, $a \in F$.
($\underline{backwards}$) Assume we have a set $F \subseteq \mathbb{R}$ such that every Cauchy sequence in F has a limit that is also in F. Then, by Theorem 2.6.2, these sequences are convergent. Now, by Theorem 3.2.5, each of these limits are actually a limit point of F. Since all of the limit points were assumed to be in F, F is closed.