Prove that,If a subset E of Real numbers is open, then complement of E is closed

Question

Our target is to show that $\forall y\in E^c \forall \epsilon >0 N'_\epsilon (y) \cap E^c \neq \phi$. Here $N'_\epsilon (y)$ means the deleted neighborhood of y. Take an arbitrary y from $E^c$. So $y \notin E$, thus y is not an interior point of E. Therefore, $\forall \epsilon>0 N_\epsilon (y) \nsubseteq E$. Now here two cases are possible. (i)$E\subseteq N_\epsilon (y) $ or (ii)$N_\epsilon (y) \cap E= \phi $. If case (i) occurs, then for all $\epsilon>0 \cap N_\epsilon (y)$ equal to E. So $y \in E$, which is a contradiction. Hence we conclude that case (ii) occurs, namely $N_\epsilon (y) \cap E= \phi $. Since $E \cup E^c =\mathbb{R}$. Therefore $ N_\epsilon (y) \subseteq E^c$. On the other hand, $N'_\epsilon (y)\subseteq N_\epsilon (y)$. So, $N'_\epsilon (y)\subseteq E^c$. Hence, $N'_\epsilon (y) \cap E^c \neq \phi$. Since y is arbitrary, so this holds for any $y\in E^c$. Therefore, every point of $E^c$ is a limit point. So, we can conclude that $E^c$ is closed. Would you kindly check my attempt? Please let me know if there is anything wrong. Thank and gratitude in advance.