Proving the derived set $E'$ is closed.

Without any separation axioms, the set of limit points of a set need not be closed. Consider a two-element set, $X = \{a,b\}$, with the trivial topology. Let $E = \{a\}$. Then $E' = \{b\}$ is not closed.

If we assume $T_1$, the set of limit points is closed. If $E' = \varnothing$, there's nothing to show, so let us suppose $E'\neq\varnothing$, and let $x \in \overline{E'}$. Let $U$ be an open neighbourhood of $x$. Then $U\cap E' \neq\varnothing$. Let $y\in U\cap E'$. If $y = x$, we're done. Otherwise, $U\setminus \{x\}$ is an open neighbourhood of $y$, and thus $(U\setminus\{x,y\}) \cap E \neq \varnothing$, since $y \in E'$. But that implies $(U\setminus \{x\}) \cap E \neq \varnothing$, and since $U$ was arbitrary, $x\in E'$.