Proof Verification of Exercise 2.9 Baby Rudin: Prove that the complement of $E^{o}$ is the closure of the complement of $E$. I am stuck on part d of this exercise:

Let $E^{o}$ denote the set of all interior points of a set. Prove that the complement of $E^{o}$ is the closure of the complement of $E$.

I solved it using double inclusion:

Let $p$ $\in (E^{o})^{c}$. Since $p$ is not an interior point of $E$, every neighborhood of $p$ contains a point not in $E$ (or else we have a neighborhood of $p$ that is contained in $E$ and p is an interior point) so $p$ is a limit point of $E^{c}$ so $p$ $\in$ $\overline{E^{c}}$. Conversely, let $q$ $\in$ $\overline{E^{c}}$. Then $q$ belongs to $E^{c}$ and is not an interior point of E, or $q$ is a limit point of $E^{c}$ and no neighborhood of $q$ is contained in $E$, so $q$ is not an interior point of $E$ and so $q \in (E^{o})^{c})$

Is this proof correct? If not, where exactly does it break down? Thank you for your time.
 A: The only thing that I would add to that is a justification why is it that, if $q$ is a limit point of $E^\complement$, then no neighborhood of $q$ is contained in $E$. That's easy, of course: since $q$ is a limit point of $E^\complement$, every neighborhood of $q$ intersects $E^\complement$, and therefore no neighborhood of $q$ is contained in $E$.
A: Here it is another way to approach it for the sake of curiosity.
Definitions
Let $(X,d_{X})$ be a metric space and $E\subseteq X$. We say that $x_{0}\in X$ is an interior point of $E$ iff there exists an open ball $B(x_{0},\delta)\subseteq E$. We also say that $x_{0}\in X$ is an exterior point of $E$ iff there exists an open ball s.t. $B(x_{0},\varepsilon)\cap E = \varnothing$. Finally, we say that $x_{0}$ is a boundary point of $E$ iff $x_{0}$ is neither an interior point nor an exterior point of $E$.
Based on such definitions, we can conclude that the interior, the exterior and the boundary of $E$ are pairwise disjoint subsets of $X$.
Hence we conclude that $\{\text{int}(E),\partial{E},\text{ext}(E)\}$ is a partition of $X$.
Proposition
Let $A\subseteq X$ and $B\subseteq X$. If $A\cap B = \varnothing$, then $A\subseteq B^{c}$.
Proof
Suppose otherwise that $A\not\subseteq B^{c}$. Then we can conclude there exists $a\in A$ such that $a\not\in B^{c}$, that is to say, $a\in A\cap B$, which contradicts our assumption. Consequently, the proposed claim is true.
Lemma
Given $E\subseteq X$, then $\text{ext}(E) = \text{int}(E^{c})$ and $\partial E = \partial E^{c}$.
Proof
Given a point $x_{0}\in X$, we say that it is an exterior point of $E$ iff there exists an open ball $B(x_{0},\delta)$ s.t.
\begin{align*}
B(x_{0},\delta)\cap E = \varnothing & \Longleftrightarrow B(x_{0},\delta)\subseteq E^{c}\\\\
& \Longleftrightarrow x_{0}\in\text{int}(E^{c})
\end{align*}
On the other hand, we say that $x_{0}\in\partial E$ iff $x_{0}$ is neither an interior point of $E$ nor an exterior point of $E$.
This means that, for every $\delta > 0$, $B(x_{0},\delta)\not\subseteq E$ and $B(x_{0},\delta)\cap E\neq \varnothing$.
This is the same as claiming that, for every $\delta > 0$, $B(x_{0},\delta)\cap E^{c}\neq\varnothing$ and $B(x_{0},\delta)\not\subseteq E^{c}$.
Thus $x_{0}\in X$ is a boundary point of $E$ iff $x_{0}$ is a boundary point of $E^{c}$.
Solution
In accordance to this answer as well as the previous results, we have that
\begin{align*}
X - \text{int}(E) & = \text{int}(E)\cup\partial{E}\cup\text{ext}(E) - \text{int}(E)\\\\
& = (\text{int}(E)\cap\text{int}(E)^{c})\cup(\partial E\cap\text{int}(E)^{c})\cup(\text{ext}(E)\cap\text{int}(E)^{c})\\\\
& = (\partial E\cap\text{int}(E)^{c})\cup(\text{ext}(E)\cap\text{int}(E)^{c})\\\\
& = \partial E\cup\text{ext}(E)\\\\
& = \partial E^{c}\cup\text{int}(E^{c})\\\\
& = \overline{E^{c}}
\end{align*}
A: Another proof :

*

*Let's prove that $E\backslash \overset{\circ}{A}\subset\overline{E\backslash A}$, which is equivalent to proving $E\backslash \overline{E\backslash A} \subset \overset{\circ}{A}$. Since $E\backslash \overline{E\backslash A}$ is an open set (the complementary of a closed set), and since it is included in $A$ (because $E\backslash A \subset \overline{E\backslash A}$), we get $E\backslash \overline{E\backslash A} \subset \overset{\circ}{A}$ because $\overset{\circ}{A}$ is the biggest open set included in $A$.

*Let's prove that $\overline{E\backslash A}\subset E\backslash \overset{\circ}{A}$. Itr is true because $\overline{E\backslash A}$ is a closed set that contains $E\backslash A$ (because $\overset{\circ} A\subset A$) and $\overline{E\backslash A}$ is the smallest closed set that contains $A$.

