Compact ∧ Connected ∧ LOTS ∧ ¬Empty ⇒ Fixed point property

Let $f\colon X\to X$ be continuous. For $x_0\in X$, if $f(x_0)>x_0$, then let $U,V$ be disjoint open neighborhoods with $x_0\in U$, $f(x_0)\in V$, and $u<v$ for each $u\in U$, $v\in V$. Then for $x\in f^{-1}(V)\cap U$, $f(x)>x$, so $A:=\{x\in X\mid f(x)>x\}$ is open, as is $B:=\{x\in X\mid f(x)<x\}$.

If $f$ has no fixed point, then $X=A\cup B$, and since $X$ is Connected, $A$ or $B$ must be empty. But if $X=A$, then $X$ has no maximum element, and likewise if $X=B$, it has no minimum element.

This contradicts that if $X$ is LOTS and Compact and not Empty, then it has a maximum and minimum element, by the remark in Math StackExchange 4786659.