Show that there does not exist a strictly increasing function $f : \mathbb Q \to \mathbb R$ such that $f(\mathbb Q) = \mathbb R$. The following exercise in an analysis text and I am trying to solve it without concepts of general topology but fail. 

Show that there does not exist a strictly increasing function $f : \mathbb Q \to \mathbb R$ such that $f(\mathbb Q) = \mathbb R$. 

Attempt 1. Suppose that the function $f(D) = \mathbb R$ is monotone. If its image $f(D)$ is an interval, then the function $f$ is continuous. So, if we suppose by contradiction that a strictly increasing function $f : \mathbb Q \to \mathbb R$ exists such that $f(\mathbb Q) = \mathbb R$ it must be continuous. 
Attempt 2. Since intersection of the set of irrational numbers of the domain is empty so by a convergence of a sequence in the domain $\mathbb Q$ in either case of converging to a rational or irrational number there is nothing to reach a contradiction. 
Attempt 3. The function $f$ is injective and it's not surjective so the inverse function is not defined such that I can use theorems about an inverse of a function. 
Please help!         

Here is a proof that does not use the uncountability of $\mathbb{R}$.  Suppose $f:\mathbb{Q}\to\mathbb{R}$ is strictly increasing, and let $\alpha\in\mathbb{R}$ be any irrational number.  Let $A=(-\infty,\alpha)\cap\mathbb{Q}$ and $B=(\alpha,\infty)\cap\mathbb{Q}$.  Then every element of $f(A)$ is less than every element of $f(B)$, so $\sup f(A)\leq \inf f(B)$.  Moreover, if $x\in A$, then $x+\epsilon\in A$ for sufficiently small rational $\epsilon>0$, so $f(x)<f(x+\epsilon)\leq \sup f(A)$.  Similarly, if $x\in B$, then $f(x)>\inf f(B)$. Thus $$f(\mathbb{Q})=f(A)\cup f(B)\subseteq (-\infty,\sup f(A))\cup (\inf f(B),\infty),$$ so any $y\in [\sup f(A),\inf f(B)]$ is not in the image of $f$.
