Scheduling: Existence of specific total order
I think you mean partial order, as if you only have a preorder then there might not even exist any total order that is compatible with it.



Your argument seems to be missing something about how you construct your topological sort, as it implies that any topological sort works, which you disproved yourself. In your example $x,y,x',y'$ could be a topological sort.

Here is an other attempt at a proof:



Let $\leq$ denote a partial order on $X$. I claim there exists a total order $\leq'$ compatible with $\leq$ such that the following never occurs:
$a\leq'\alpha\leq'b$
where $a \leq b$, and $\alpha$ is incomparable to both $a$ and $b$. 
(Note that when I say comparable/incomparable I always mean with respect to $\leq$)

To see this choose a compatible total order $\leq'$ with the least amount of such "bad" pairs $(a\leq b)$. Suppose that this amount is non-zero and let $(a\leq b)$ be such a pair with smallest $a$ (with respect to $\leq'$). Let $\alpha_1 \leq' \alpha_2 \leq'\ldots\leq'\alpha_k$ denote all elements between $a$ and $b$ which are incomparable to both $a$ and $b$.

Now create a new total order $\leq^*$ from $\leq'$ by moving all $\alpha_i$'s right in front of $a$ to get $\alpha_1 \leq^* \alpha_2 \leq^*\ldots\leq^*\alpha_k\leq^*a$, the rest remaining unchanged. Notice that this is still compatible with $\leq$, as the only thing that could have gone wrong is that there was some $a\leq c \leq \alpha_i$ but this is impossible as $a$ and $\alpha_i$ are incomparable.

Notice also that, because all elements between $a$ and $b$ (with respect to $\leq^*$) are now comparable to $b$, we have created no "bad" pair $(\alpha_i\leq x)$ with $a\leq^* x \leq^* b$, otherwise we would have $\alpha_i\leq x \leq b$ meaning that $\alpha_i$ and $b$ are comparable. And we have created no "bad" pair $(x\leq a)$, otherwise $(x\leq b)$ would have been a bad pair to start with, contradicting our choice of $a$. Any other type of "bad" pair we created must have already existed previously.

Because $(a,b)$ is no longer a bad pair, $\leq^*$ has strictly less bad pairs than $\leq'$, which is impossible by definition of $\leq'$. Thus, $\leq'$ had no bad pairs to start with. 

It is easy to show from here that $\leq'$ meets your requirements.