Proving that the closure of a set is closed directly

Currently working through Rudin's principle's of mathematical analysis. I am trying to prove directly that the closure of a set is closed but am hitting a wall on one part of the proof. Namely, if we take the closure of a set to be the union of the set itself with the set of its limit points. We can say:

Let $x$ be a limit point of $\bar{E}$. then $\forall r>0,\ N_r(x) \cap \bar{E}$ is nonempty. By definition of $\bar{E}, (N_r(x) \cap E) \cup (N_r(x) \cap E')$ is nonempty. This implies at most one of these is empty.

1. if $N_r(x) \cap E$ is nonempty, $x$ is limit point of $E \implies x \in E' \implies x \in \bar{E}$
2. if $N_r(x) \cap E'$ is nonempty, $x$ is limit point of $E' \implies ....$

2. is where I'm not sure how to proceed.