This is exercise 15G.3 in Willard's General Topology, 1970. The writer has a suggestion: Suppose $x_n\rightarrow p$ but $x_n$ is not ultimately constant. Construct sequence $U_1, U_2,...$ of disjoint open sets in $X$ such that $x_{n_{k}} \in U_k$ for some subsequence $x_{n_{k}}$ of $x_n$ and such that $p\in \overline{U_k}$ for every $k$. Let $G=\cup _{k=1}^{\infty} U_{2k}$. Then $\overline{G}$ is open containing $p$ but $x_{n_{k}}$ dont belong to $\overline{G}$ for odd $k$. This is the contradiction.
I remark here the suggestion is wrong :since the space is extremally disconnected the sets $\overline{U_k}$ are also disjoint (15G.1), so it's not possible $p\in \overline{U_k}$ for every $k$. My first question: is there something i cannot see here and my comment is wrong?
My second question is to check please the following proof i provide:
Suppose $x_n\rightarrow p$ but $x_n$ is not ultimately constant. Using T2 we construct sequence $U_1, U_2,...$ of disjoint open sets in $X$ such that $x_{n_{k}} \in U_k$ for some subsequence $x_{n_{k}}$ of $x_n$. We set $G=\cup_{k=1}^{\infty} \overline{U_k}$ which is open, since $\overline{U_k}$ are open. We consider the function $f:G\rightarrow R$ with $f(\overline{U_k})=k$, $k=1,2,...$ It is continuous on $G$: Let $ε>0$ small enough and $(k-ε,k+ε)$ an open nbhood of $k$ in $R$. Then $f^{-1}(k-ε,k+ε)=f^{-1}(k)=\overline{U_k}$ open. But the space $G$ is open in $X$, so by 15G.1, it is $C^{*}$-embedded to $X$, see 15E. That gives there exists a continuous extension $F:X\rightarrow R$, $F|_G=f$. By continuity, $F(x_{n_{k}})\rightarrow F(p)\in R$. But $F(x_{n_{k}})=k\rightarrow +\infty$. We arrived at a contradiction.
