Prove that $\mathbb{R}^2\setminus E$ is path-connected 
Let $E$ be the set of all points in $\mathbb{R}^2$ having both coordinates rational. Prove that the space $\mathbb{R}^2\setminus E$ is path-connected.

Path-connected definition: A topological space $(X,\tau)$ is said to be path-connected if given $a,b\in X$, there exists a continuous function $f:[0,1]\to X$ such that $f(0)=a$ and $f(1)=b$. 
I have read a similar thread on mathstackexchange but I am failing to build the function that proves that any two points of $\mathbb{R}^2\setminus E$ are path-connected.
If we consider $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2\setminus E$ so that $x_1,y_2$ are irrational as proposed in the answer of another question.
I can build two functions $f:(x_1,y_1)\to(x_1,y_2)\\(x_1,y_1)\to(x_1,y_1+c)$ 
so that $c\in\mathbb{R}$
$g:(x_1,y_2)\to(x_2,y_2)\\(x_1,y_1)\to(x_1+d,y_2)$ so that $d\in\mathbb{R}$
So $f \circ g:(x_1,y_1)\to(x_2,y_2)$.
However this is not a generalization for all the points in $\mathbb{R}^2\setminus E$ and I cannot relate the function to the interval $[0,1]$.
Question:
How should I solve the exercise?
Thanks in advance!

Let $(a,b), (a',b') \in \mathbb Q^2$  be two points of $\mathbb R^2$ with both rational coordinates. Consider two sequences $(a_n)_{n\in \mathbb Z},(b_n)_{n\in \mathbb Z}$ of irrational numbers such that 
$$ \lim_{-\infty}a_n = a\quad \lim_{+\infty} a_n=a', \quad \lim_{-\infty} b_n=b,\quad \lim_{+\infty} b_n=b',$$.
Now consider the $\mathbb Z-$sequence of points 
$$\left\{\begin{matrix}x_{2n+1}&=& (a_n+1,b_n) \\x_{2n}&=&(a_n,b_{n}) \end{matrix}\right.$$
and notice that  $$\forall n\in \mathbb Z, \forall t\in [0,1],~~ (1-t)x_n+tx_{n+1} \notin E$$
you then construct a piecewise linear path $\gamma: [-1,1]\rightarrow \mathbb R^2$ such that $\forall n\in \mathbb Z, \gamma(\tanh(n))=x_n$. 
By definition, $\lim_{t\rightarrow -1}\gamma(t)=(a,b)$
 and $\lim_{t\rightarrow 1} \gamma(t)=(a',b')$
