Problem:Let $\displaystyle E'$ be the set of all limit points of $\displaystyle E \subset X$. Prove that $\displaystyle E'$ is closed.

I think I can do it if the set $\displaystyle E'$ has the least-upper-bound property.

Attempt:Let $\displaystyle p$ be a limit point of $\displaystyle E'$. (If the set of $\displaystyle p$ is finite or empty then $\displaystyle E'$ is closed.) Then for every $\displaystyle r > 0$ there exists a point $\displaystyle q \ne p$ such that $\displaystyle q \in E'$. Let $\displaystyle A$ be the set of points of $\displaystyle q \in E'$ such that $\displaystyle q<p$. This is bounded above by $\displaystyle p$, which is the lub of $\displaystyle E'$. Hence $\displaystyle \text{sup}(A) = p$ and $\displaystyle p \in E'$ by the least-upper-bound property. Since $\displaystyle p$ was arbitrary, this proves that $\displaystyle E'$ is closed.

Firstly, does this sound okay? If so, how could I prove this without using the least-upper-bound property?

I don't have anybody to look at my work, so I may be making huge fallacies or assumptions, and I'd greatly appreciate any input on this. Thanks.