Let $X$ and $Y$ be topological spaces and let $\succeq$ be a total preorder (a complete and transitive binary relation) on $X\times Y$. Assume that $\succeq$ is continuous, that is, $\succeq$ is a closed subset of $X\times Y\times X\times Y$ with respect to the product topology. Assume further that there exists a continuous surjective function $u:X \rightarrow \mathbb{R}$ (where $\mathbb{R}$ is equipped with the usual (=order) topology) such that $(x,y) \succeq (x',y)$ $\Leftrightarrow$ $u(x) \geq u(x')$.
Is the total preorder $\succeq'$ on $\mathbb{R}\times Y$ defined by $(u(x),y) \succeq' (u(x'),y')$ $\Leftrightarrow$ $(x,y) \succeq (x',y')$ continuous (i.e. $\succeq'$ is a closed subset of $\mathbb{R}\times Y\times \mathbb{R}\times Y$)?
UPD. My attempt: I think of the quotient topology on $X\times Y/ \approx$, where the equivalence relation $\approx$ is defined as $(x,y) \approx (x',y')$ $\Leftrightarrow$ $u(x)=u(x')$ & $y=y'$. However, I have failed to prove that $X\times Y/ \approx$ is homeomorphic to $\mathbb{R}\times Y$.
The statement seems to be false. A counterexample: $X=\mathbb{R}$ with the discrete topology, $Y=\mathbb{R}$ with the usual topology, $\succeq$ is the lexicographical order on $\mathbb{R}\times\mathbb{R}$, and $u$ is the identity function.