Is there a nondiscrete linearly ordered topological space (LOTS) that is extremally disconnected?

Question

A space is extremally disconnected if the closure of every open set is open. I was wondering if there is a ordered space that is extremally disconnected but not discrete.

Currently there is no such example in $\pi$-base. Note that by this question, such space must be sequentially discrete, and hence non-sequential. Also, changing LOTS to the weaker condition $T_5$ results in a single example called the single ultrafilter topology. Any help appreciated for constructing/disproving the existence of such a space.

Edit. Inspired by the existing answer I can now construct a sequentially discrete LOTS that is not discrete:

Consider $X = (\omega_1\times\mathbb{Z})\cup\{\infty\}$, with lexicographical order on $\omega_1\times\mathbb{Z}$ and that $\infty$ is larger than any other element. Every point other than $\infty$ is an isolated point because $\omega_1\times\mathbb{Z}$ is discrete, but $\infty$ is not isolated since there is no largest element in $\omega_1\times\mathbb{Z}$. On the other hand, $\{\infty\}$ is sequentially open, which is to say there is no sequence in $\omega_1\times\mathbb{Z}$ converging to $\infty$: For every $((a_n,b_n))_{n\in\omega}\in\omega_1\times\mathbb{Z}$, the sequence $(a_n)_{n\in\omega}$ is bounded above by some $\alpha\in\omega_1$, so $((a_n,b_n))_{n\in\omega}$ is bounded above by $(\alpha+1,0)$.

Edit 2. The proof in the answer generalizes easily to GO-spaces (generalized ordered spaces). A GO-space is a topological space $X$ with topology $\tau$ together with a linear order $<$ such that $X$ is $T_1$ and every point has a local base of $\tau$-open neighborhoods consisting of order-convex sets.

Suppose that $x$ is not an isolated point, then $(\leftarrow,x)\cup(x,\rightarrow)$ is dense, so $x\in\overline{(\leftarrow,x)}$ or $x\in\overline{(x,\rightarrow)}$, then $x$ lies in the closure of every cofinal subset of $(\leftarrow,x)$ or $(x,\rightarrow)$ by order-convexity of local base elements of $x$: If $U$ is an order-convex open neighborhood of $x$, then $y\in U\cap(\leftarrow,x)$ (resp. $y\in U\cap(x,\rightarrow)$) implies that $U$ contains every element $\ge y$ (resp. $\le y$) of the cofinal subset. The rest of the proof remains the same.

Answer 1

No, there is no such space. The argument is a slight generalisation of the observation that $\omega + 1$ is not such a space because the closure of $\{2, 4, 6, 8, \dotsc\}$ is not open.

Suppose $(X, \le)$ is a total order which is not discrete in the order topology. This means there is some $x \in X$ such that $\{x\}$ is not open. From this it follows that either $(-\infty, x)$ is non-empty and has no largest element, or $(x, \infty)$ is non-empty and has no smallest element. (Otherwise, $x$ would have a predecessor and a successor in $X \cup \{\pm \infty\}$ and hence $\{x\}$ could be written as an open interval).

Without loss of generality, suppose $(-\infty, x)$ is non-empty and has no largest element. Since any total order has a well-ordered cofinal subset, we may assume there is a limit ordinal $\lambda$ and a strictly increasing transfinite sequence $(y_\alpha : \alpha < \lambda)$ which is cofinal in $(-\infty, x)$.

Now consider the set $U = (y_0, y_2) \cup (y_4, y_6) \cup (y_8, y_{10}) \cup \dotsb \cup (y_\omega, y_{\omega + 2}) \cup \dotsb$ (formally, $U = \bigcup_{\beta < \lambda\text{ a limit}} \bigcup_{n < \omega} (y_{\beta + 4n}, y_{\beta + 4n + 2})$).

Then $U$ is open, and by the cofinality of the $y_\alpha$, the closure of $U$ contains $x$. But also, the closure of $U$ does not contain $y_3, y_7, y_{11}, \dotsc, y_{\omega + 3}, \dotsc$, which is a cofinal sequence in $(-\infty, x)$. So the closure is not open, because any neighbourhood of $x$ must contain some terminal segment of $(-\infty, x)$.