I was reading the proof in Rudin, but it uses the metric. Is this not true if $X$ is a general topological space and $E' \subset X$ (especially if it is not Hausdorff $T_1$)?
I can't come up with a counterexample. To be specific, it's difficult for me to pass from one neighbourhood to the other without using a metric.
Note that $E'$ is the set of all limit points.
Best Answer
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, $x\in E'$.