I'm aware that it is duplicate, but I'd like to know whether my example is appropriate or not.
Let our function $f$ be on the set $\mathbb{Q}\cap\mathbb{Z}$ induced by standard topology of a line. I'm going to construct a bijection from $\mathbb{Q}\cap\left((0,1)\cup(1,2)\right)$ to $\mathbb{Q}\cap(0,1)$. The rest from $(2n,2n+1)\cup(2n+1,2n+2)$ to $(n,n+1)$ is similar. My idea is that if the restriction of $f$ on $(n,n+1)$ is monotonous then we are done, because $f$ becomes continuous but its inverse does not. Let's denote elements from $(0,1)$ in domain as $\dfrac{a'}{b}$ and elements from $(1,2)$ in domain decrease by $1$ and denote as $\dfrac{a''}{b}$ . $$f(\dfrac{1'}{2})=\dfrac{1}{2}\text{and}f(\dfrac{1''}{2})=\dfrac{1}{3}$$ $$f(\dfrac{1'}{3})=\dfrac{1}{4}\text{and}f(\dfrac{1''}{3})=\dfrac{1}{5}$$ $$f(\dfrac{2'}{3})=\dfrac{2}{3}\text{and}f(\dfrac{2''}{3})=\dfrac{3}{4}$$ $$f(\dfrac{1'}{4})=\dfrac{1}{6}\text{and}f(\dfrac{1''}{4})=\dfrac{1}{7}$$ $$...\text{so.on}...$$ So $f(x'')=y$ means $f(1+x)=y$ and $f(x')=f(x)$
The idea is next: We start from the elements with smallest denominator and map them on element which satisfies the condition of monotone on the one hand, and has the smallest possible denominator on the other hand. I think that this idea will lead us to surjectivity, but still I doubt.
P.S. There is an assumption that for $\dfrac{p}{q}$ we have $\gcd(p,q)=1$