The basic reason that there can be no monotone mapping of $\mathbb{R}$ onto $\mathbb{R}\setminus\mathbb{Q}$ is the completeness of the linear order on $\mathbb{R}$.
Suppose that $f:\mathbb{R}\to\mathbb{R}\setminus\mathbb{Q}$ is an increasing surjection. Let $L=\{x\in\mathbb{R}:f(x)<0\}$; clearly $L$ is bounded above (e.g., by the real number $y$ such that $f(y)=\sqrt2$), so $L$ has a least upper bound $u$. Now what can $f(u)$ be?
If $u\notin L$, then $f(u)>0$, and $f(x)\ge f(u) > 0$ for every $x\ge u$, so $f[\mathbb{R}]\cap(0,f(u))=\varnothing$, and $f$ isn’t a surjection.
If $u\in L$, then $f(u)<0$, but $f(x)>0$ for every $x>u$, so $f[\mathbb{R}]\cap (f(u),0)=\varnothing$, and again $f$ is not a surjection.
In either case we have a contradiction, so $f$ cannot be a surjection.
If $f$ were a decreasing surjection, $-f$ would be an increasing surjection, so a decreasing function from $\mathbb{R}$ to $\mathbb{R}\setminus\mathbb{Q}$ can’t be a surjection either.
If you're going to deal first with monotone increasing functions, you should begin like this:
Assume that $f$ is monotone increasing on $[a,b]$ and discontinuous at $x_0\in[a,b]$.
Now you should use the definition of continuity at a point to see what this tells you about $f$.
We know that $f(x_0)$ exists, so it must be the case that $$\lim_{x\to x_0}f(x)\ne f(x_0)\;.$$
That pretty well exhausts what we can say just on the basis of the assumption that $f$ is discontinuous at $x_0$, so let's look at what the monotonicity of $f$ can tell us. As $x$ approaches $x_0$ from the left, $f(x)$ is increasing; does it have a limit? Yes, because of the completeness of $\Bbb R$. But if we're going to look at $\lim\limits_{x\to x_0^-}f(x)$, we'd better not let $x_0=a$. In fact, the cases $x_0=a$ and $x_0=b$ both look as if they might require a little special handling, so let's set them aside for the moment.
Assume for now that $a<x_0<b$. Since $f$ is increasing, $f(x)\le f(x_0)$ whenever $a\le x<x_0$, so $\{f(x):a\le x<x_0\}$ is bounded above by $f(x_0)$ and therefore has a least upper bound, say $L$.
Now prove that $$L=\lim_{x\to x_0^-}f(x)\;.$$ Go on to prove in similar fashion that $\lim\limits_{x\to x_0^+}f(x)$ exists, and call it $R$, say.
The final step is to ask yourself how $\lim\limits_{x\to x_0}f(x)$ can fail to exist when the one-sided limits $L$ and $R$ both exist?
Then you have to clean up the loose ends when $x_0$ is $a$ or $b$, but that's easy once you have the main part done. You also have to deal with monotone decreasing $f$. You can either repeat the argument above with very minor changes, or you can look at $-f$: if $f$ is decreasing, then $-f$ is increasing, so you already know that it has only jump discontinuities, and from that you should be able to show very quickly that the same is true of $f$.
Best Answer
At each jump, draw an open interval on the $y$-axis which fits in between the bottom of the jump and the top of the jump.
After you are done, you have a string of disjoint intervals on the $y$-axis (because the function is increasing.)
Now choose a rational number in each of these intervals. This gives you a bijection between the intervals and a subset of $\Bbb Q$.
The main idea is that there is a limit to how big a disjoint collection of open sets in $\Bbb R$ can be. The generalized idea is that of the Souslin number of a topological space. In a nutshell, it's "the largest cardinality of a set of disjoint open sets."
You can see that the same argument works with $\Bbb R$ replaced with any separable linearly ordered topological space.