Let $A=\{e: \varphi_e$ is total and $\forall x(\varphi_e(x)\leq\varphi_e(x+1))\}$, where $\varphi_e$ denotes the $e$-th recursive function.
I'm trying to show that the previous set is recursively enumerable, since I've already proved that it is not recursive. My attempt has been to express it as an intersection of $B=\{e: \varphi_e$ is total$\}$ and $C=\{e:\forall x(\varphi_e(x)\leq\varphi_e(x+1))\}$. I know that $B$ is recursively enumerable, and I think $C$ is as well, but I don't know how to show it, because as far as I know $\forall x P(x)$ is not a recursive predicate in general even when $P(x)$ is recursive.
$B$ is not recursively enumerable. While for any fixed $k$, we can wait and see whether $\varphi_e(k)$ halts - and so for any fixed $k$, the set $\{e: \varphi_e(k)$ halts$\}$ is r.e. - this fails when we consider infinitely many $k$ at once.
The intuition here is just to think about the task at hand: if I'm playing around with a Turing machine $\varphi_e$, how can I be sure that it's total after examining it for only finitely much time? At best I can see $\varphi_e$ halt on finitely many inputs, but it could still fail to halt on others.
In fact, $B$ has Turing degree $0''$ - strictly more complicated than the halting problem.
In fact, you can reduce $B$ to the set you're interested in (and so it's not r.e.): do you see a way to transform a computable partial function $f$ into a computable partial function $\hat{f}$ such that $f$ is total iff $\hat{f}$ is total and nondecreasing?