I need help understanding the following proof:
Prove that $\Bbb Q$ is dense in $\Bbb R$, i.e.
$$\forall \varepsilon>0,\forall x\in\Bbb R, \ (x-\varepsilon,x+\varepsilon)\land \Bbb Q\neq \emptyset.$$
Let's assume the opposite:
$\exists\varepsilon>0, \exists x\in\Bbb R, \forall q\in\Bbb Q, (q\leq x-\varepsilon)\lor (x+\varepsilon\leq q)$.
Let $q_1\leq x-\varepsilon$ and $x+\varepsilon\leq q_2$. Let $n\in\Bbb N$ such that $2n\varepsilon>q_2-q_1$ and let $\alpha_k=q_1+\frac{k}{n}(q_2-q_1)\in\Bbb Q (k=0,1,...,n-1,n)$. Let $1\leq k_0<n$ be the greatest index for which $\alpha_k\leq x-\varepsilon$. Now because of $x+\varepsilon\leq\alpha_{k_0+1}$ we have $\frac{q_2-q_1}{n}=\alpha_{k_0+1}-\alpha_{k_0}\geq 2\varepsilon$ which is a contradiction with the choice of $n$.
I'm completely lost. Can someone tell me what's the idea behind this proof?
The idea is that if you have a hole of diameter $2\epsilon$ and a frog whose jumping length is $\frac{1}{n}$ (rational) and less than the diameter of the hole $\left(\frac{1}{n}<2\epsilon\right)$, and the frog is initially located in origin and starts jumping in the direction of the hole, it will eventually end up in the hole.