I'm trying to prove that. Let $(X,\leq)$ a partial ordered set, with the associated topology formed by the subsets $U$:
$ x \in U$ and $ y \leq x \Rightarrow y \in U$
Then a function $f$ is continuous if and only if it preserves the order, i.e, $x\leq y \Rightarrow f(x)\leq f(y)$.
Any help?
Let $X$ and $Y$ be partially ordered sets and $f\colon X\to Y$ a map.
First, assume that $f$ is increasing, and let $U$ be an open set in $Y$. Let $x\in f^{-1}(U)$ and let $y\in X$ such that $y\leq x$. As $f$ is increasing, then $f(y)\leq f(x)$, thus $f(y)\in U$ by definition of the open sets in $Y$. Then $y\in f^{-1}(U)$ and this shows that $f^{-1}(U)$ is open in $X$. Hence, $f$ is continuous.
Conversely, assume that $f$ is continuous and let $x,y\in X$ such that $y\leq x$. Let $$ U= \{z\in Y \ \colon \ z\leq f(x)\}. $$ This is an open set in $Y$ because if $z\in U$ and $w\leq z$ then $w\leq z\leq f(x)$ and $w\in U$. Thus $U$ is an open neighborhood of $f(x)$ in $Y$ and by continuity, $f^{-1}(U)$ is an open neighorhood of $x$ in $X$. As $y\leq x$ we have that $y\in f^{-1}(U)$ and consequently, $f(y)\in U$. By definition of $U$ this means that $f(y)\leq f(x)$, which shows that $f$ is increasing.
Remark. We don't need the sets $X$ and $Y$ to be partially ordered but only preordered. That is, the relation $\leq$ only needs to be refletive and transitive, but not necessarily antisymmetric.