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If $E^{\circ}$ denotes the set of interior points of $E$, prove $E^{\circ}$ is open.

My attempt: If $x\in (E^{\circ})^c$ then $x$ is not an interior point of $E$. Therefore, no neighborhood of $x$ is contained entirely in $E$ and thus all neighborhoods of $x$ contain a point $p\in E^c \subset (E^{\circ})^c$, hence $x$ is a limit point of $(E^{\circ})^c$ therefore $(E^{\circ})^c$ is closed and $E^{\circ}$ is open.

Is there something wrong with this proof? I ask because all the proofs I have seen of this fact avoid using this argument, which somehow raises doubts about its validity, given it is just as simple as other arguments used.

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    Just to be clear, are you talking about metric spaces? – 2017-01-05
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    @Aweygan yes I am. – 2017-01-05
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    Every closed set $A$ is $E^{\circ c}$ for some $E$ (let $E=A^c$). So you proved that $x\in A \implies x$ is a limit point of $A$ for every closed set $A$. This is clearly false. For example, a singleton in any discrete space. – 2017-01-05

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Your argument is perfectly fine but is more cumbersum. The openness of $E^\circ$ can be proved directly without invoking closedness:

Since $E^\circ$ is a neighbourhood (contains a ball at) of all of its points, $E^\circ$ is open.

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    sorry, how does this show that $E^{\circ}$ is open? Yes, each point in $E^{\circ}$ has a neighborhood contained in $E$, but you would need to show that the neighborhood is contained in $E^{\circ}$, correct? – 2017-01-06
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    @DanielXiang That follows immediately from triangle inequality. Let $B:= \operatorname{Ball}(x;r) \subseteq E$, then if $y \in B$, $\operatorname{Ball}(y;r-d(x,y)) \subseteq E$, so $B \subseteq E^\circ$. – 2017-01-06
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    ah, I see now. Since every point in $B(x,r)$ is an interior point of $E$, we must have $B(x,r) \subset E^{\circ}$. Thank you! – 2017-01-06