When does $E(x|s=\omega)$ increase in $\omega$? Let $x\sim U[0,1]$ be the "truth", which we observe through an imperfect signal $s$, such that:
$$s=x+\epsilon, \text{ where } E(\epsilon|x)=0.$$
Moreover, the $\epsilon$ is such that $s$ is still in $[0,1]$ (thus, $\epsilon$ is not independent of $x$). This is a strong and crucial assumption (thanks for pointing it out!)
I would like to show (or determine under what additional conditions it would hold) that the conditional expectation of the true $x$ given the signal realization is increasing in the signal realization, i.e.
$$E(x|s=\omega)\text{ is increasing in }\omega.$$
It seems intuitive, yet I have no idea how to prove it.
 A: You claim that
$$E(x|s=\omega)\text{ is increasing in }\omega.$$
This is false without any further assumptions. For example, let $\epsilon$ be determined as follows:
Flip a coin (independent of $x$), if heads, set $\epsilon=-2x$, and if tails, set $\epsilon=+2x$.
It is clear that the expectation of $\epsilon$ is zero. However, if $x=c$,
then $s\in\{-c,3c\}$. Moreover, once $s$ has been observed, one deduces $x$, it is
$$x=\begin{cases}-s&\text{if }s<0\\ s/3&\text{if }s\geq 0\end{cases}.$$
Therefore 
$$E(x|s=\omega)=-s$$
for $\omega<0$, which is decreasing.
Even if we assume the random variable $\epsilon$ to be independent of $x$, it is possible to construct counterexamples. For example, let $\epsilon$ be either $-1/2$ or $1/2$, determined also by a coin flip independent of $x$. Then again one determines
$$x=\begin{cases}s+1/2&\text{if }s<1/2\\ s-1/2&\text{if }s\geq 1/2\end{cases}.$$
This is not an increasing function, as there is a jump down at $s=1/2$.
If you choose $\epsilon$ to be a normal Gaussian independent of $x$ then you are right, which would follow from explicitly calculating
$$E(x|s=\omega).$$
PS (hope I am not being rude here) The other answer contains a mistake, I cannot comment on it because lack of reputation. Suppose that $\epsilon$ has a normal distribution. Then we can observe $s=\omega>1$. The other answer then implies
$$E(x|s=\omega)=\omega>1,$$
contradicting that $x$ has the uniform distribution on $[0,1]$, and is in particular not larger than $1$.
PPS a new counterexample given the new edit of the question:
If $x<1/5$ or $x>1/2$, set $\epsilon=0$.
If $1/5<x<2/5$, set $\epsilon=±1/5$, the sign determined by a coin flip.
So far we are certain that $1/5<s<2/5$ implies that $2/5<x<1/2$.
Finally if $2/5<x<1/2$, then we flip a coin, if heads then
$\epsilon = 4/5-2x$, if tails then $\epsilon = 2x-4/5$.
Then, for $1/5<s<2/5$, we are sure that the coin was heads, and we derive
$$E(x|s)=4/5-s.$$
Although what you say may seem intuitive, it is really not true in general and there are no general theories for it in probability theory that I know of. What you say crucially depends on the choice of error distributions (plural since you have to choose one for each $x$ seperately)!
