Hint. Use Hilbert's Hotel to lodge some numbers in $(0,1)$; then deal with the two newcomers $0$ and $1$.
That is: find a sequence $x_1,x_2,\ldots,x_n,\ldots$ of numbers in $(0,1)$, all distinct. Map $x_1$ to $0$; map $x_2$ to $1$. Map $x_3$ to $x_1$. Map $x_4$ to $x_2$. And so on. As for the rest of the numbers in $(0,1)$, well, they can just stay where they are...
Let $I = [0,1]\setminus \mathbb{Q}$, the set of irrationals in the unit interval.
Partition $I$ into countably many nonempty pieces, each given by the intersection of an open set in $[0,1]$ and $I$, such that $I = \cup_{k=0}^\infty I_k$. For example,
$I_0 = \left(\frac{1}{2},1\right) \cap I$,
$I_1 = \left(\frac{1}{4},\frac{1}{2}\right)\cap I$,
and in general $I_k = \left(\frac{1}{2^{k+1}},\frac{1}{2^k}\right)\cap I$. Note that each $I_k$ is open in $I$.
Enumerate $\mathbb{Q} \cap [0,1]$ (in any way you like): $\{q_0, q_1, q_2,\dots\}$.
For any $x\in I$, $x\in I_k$ for exactly one $k$. Define $f(x) = q_k$.
Note that $f$ is surjective, since each $I_k$ is nonempty.
We want to show that $f$ is continuous. Let $O\subseteq \mathbb{Q}\cap [0,1]$ be an open set. We can write $O = \bigcup_{q_k\in O} \{q_k\}$.
$$f^{-1}[O] = \bigcup_{q_k\in O}\,f^{-1}[\{q_k\}] = \bigcup_{q_k\in O} I_k.$$
This is a union of open sets in $I$, so it is open.
Note that we didn't actually use that $O$ was an open set: our function $f$ is continuous even if we give $\mathbb{Q}$ the discrete topology!
Best Answer
(1) Choose an infinite countable set of irrational numbers in $(0,1)$, call them $(r_n)_{n\geqslant0}$.
(2) Enumerate the rational numbers in $(0,1)$ as $(q_n)_{n\geqslant0}$.
(3) Define $f$ by $f(q_n)=r_{2n+1}$ for every $n\geqslant0$, $f(r_n)=r_{2n}$ for every $n\geqslant0$, and $f(x)=x$ for every irrational number $x$ which does not appear in the sequence $(r_n)_{n\geqslant0}$.
Let me suggest you take it from here and show that $f$ is a bijection between $(0,1)$ and $(0,1)\setminus\mathbb Q$.