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Suppose $p \in A \setminus E(A)$ and choose open $U$ such that $p \in U \subseteq cl(U) \subseteq X \setminus \{a,b\}$ and $\mid \operatorname{fr}_X(U) \mid \leq 2$. $U$ must have two boundary points within $A \setminus E(A)$, say $c$ and $d$. $p$ must be an ordinary point of the arc $cd$.

Suppose $q \in U \setminus A$. Cover $cd$ with open sets which have only two boundary points and which do not contain $a$, $b$ or $q$. Take a finite subcovering and call its union $V$. $V$ has only two boundary points, lying in $A \setminus cd$. So $q \in cl(U) \setminus V = U \setminus cl(V)$, which is clopen. But $X$ is connected. So $U \setminus A$ is empty and $U \subset A$. Therefore $A \setminus E(A)$ is open.

Suppose $p \in A \setminus E(A)$ and choose open $U$ such that $p \in U \subseteq cl(U) \subseteq X \setminus \{a,b\}$ and $\mid \operatorname{fr}_X(U) \mid \leq 2$. $U$ must have two boundary points within $A \setminus E(A)$, say $c$ and $d$. $p$ must be an ordinary point of the arc $cd$.

Suppose $q \in U \setminus A$. Cover $cd$ with open sets which have only two boundary points and which do not contain $a$, $b$ or $q$. (Their boundaries will all be within $A$.) Take a finite subcovering and call its union $V$. The boundary of $V$ is contained in the union of the boundaries of the members of the subcovering, and so is contained in $A \setminus cd$. So $q \in cl(U) \setminus V = U \setminus cl(V)$, which is clopen, contradicting $X$ is connected.Hence there cannot be such a $q$, $U \setminus A$ is empty and $U \subset A \setminus E(A)$. Therefore $A \setminus E(A)$ is open.

Suppose $p \in A \setminus E(A)$ and choose open $U$ such that $p \in U \subset X \setminus \{a,b\}$ and $\mid \operatorname{fr}_X(U) \mid \leq 2$. $U$ must have two boundary points within $A \setminus E(A)$, say $c$ and $d$. $p$ must be an ordinary point of the arc $cd$.

Suppose $q \in U \setminus A$. Cover $cd$ with open sets which have only two boundary points and which do not contain $a$, $b$ or $q$. Take a finite subcovering and call its union $V$. $V$ has only two boundary points, lying in $A \setminus cd$. So $q \in cl(U) \setminus V = U \setminus cl(V)$, which is clopen. But $X$ is connected. So $U \setminus A$ is empty and $U \subset A$. Therefore $A \setminus E(A)$ is open.

Suppose $p \in A \setminus E(A)$ and choose open $U$ such that $p \in U \subseteq cl(U) \subseteq X \setminus \{a,b\}$ and $\mid \operatorname{fr}_X(U) \mid \leq 2$. $U$ must have two boundary points within $A \setminus E(A)$, say $c$ and $d$. $p$ must be an ordinary point of the arc $cd$.

Suppose $q \in U \setminus A$. Cover $cd$ with open sets which have only two boundary points and which do not contain $a$, $b$ or $q$. Take a finite subcovering and call its union $V$. $V$ has only two boundary points, lying in $A \setminus cd$. So $q \in cl(U) \setminus V = U \setminus cl(V)$, which is clopen. But $X$ is connected. So $U \setminus A$ is empty and $U \subset A$. Therefore $A \setminus E(A)$ is open.