How to show that $\succ$ is open in $X^2$. 
Suppose $X$ is a topological space and $X^2$ is given the product topology. Given complete, reflexive and transitive ordering $\succeq\subset X^2$, define the relation $\succ\in X^2$ by: $(x,y)\in\succ$ if and only if $(x,y)\in\succeq$ and $(y,x)\notin \succeq$. Show that, if $\{y\in X |(y,x)\in\succeq\}$ and $\{y\in X |(x,y)\in\succeq\}$ are closed in $X$ for every $x\in X$, then $\succ$ is open in $X^2$.

$\succeq$ is reflexive if $x\in X$ implies $(x,x)\in \succeq$. $\succeq$ is  transitive if for all $x,y,z ∈ X$, $(x , y)\in \succeq$ and $(y , z)\in \succeq$ imply $(x , z)\in \succeq$. $\succeq$ is complete if for all $x,y ∈ X$, $x \ne y$ and $(x,y)\notin \succeq$ imply $(y , x)\in\succeq$.
My try: We can see that $\{y\in X |(y,x)\in\succ\}$ and $\{y\in X |(x,y)\in\succ\}$ are open in $X$ for every $x\in X$, take $(a,b)\in \succ$. Then $U(a)=\{p\in X |(p,b)\in\succ\}$ is an open neighborhood around $a$ and $V(b)=\{q\in X |(a,q)\in\succ\}$ is an open neighborhood around $b$. Let $(m,n)\in U(a)\times V(b)$, then $(m,b),(a,n)\in \succ$. But I can't prove that $(m,n)\in\succ$.
 A: Take $(a,b) \in \succ$. This means that $a \succ b$.
Case 1: If there is some $c \in X$ such that $a \succ c$ and $c \succ b$, then form $U(c) := \{x \in X: x \succ c\}$ which is open as you say, as its complement is closed by assumption. Also $a \in U(c)$.
Also $L(c) := \{x\in X: c \succ x\}$ is open (same reason) and $b \in L(c)$.
And $$(a,b) \in U(c) \times L(c) \subseteq \succ$$ whwere the last inclusion follows because $(x,y) \in U(c) \times L(c)$ means $x \succ c$ and $c \succ y$, so by transitivity $x \succ y$ as required.
Case 2: if there is no such $c$ between $a$ and $b$, then define $L(a) = \{x: a \succ x\}$ which is open and contains $b$ and $U(b) = \{x: x \succ b\}$ which is open and contains $a$. 
Now $(a,b) \in U(b) \times L(a) \subseteq \succ$: take $(x,y) \in U(b) \times L(a)$, so that $x \succ b$ and $a \succ y$. Then if $y \succeq x$ would hold, then $a \succ y \succeq x \succ b$, contradicting that we are in the case that there are no points inbetween $a$ and $b$, so we have that $y \succeq x$ does not hold, which by being a total order means that $x \succ y$, so $(x,y) \in \succ$.
