It can also give another solution, different from the oficial solution (who uses countable sets)? Without countable sets?

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@ above: the statement in the beginning does not express that $f$ is discontinuous in $x_0$ in general: discontinuity means that some $\epsilon>0$ exists such that any interval $[x_0-\delta,x_0+\delta]$ with $\delta>0$ contains an $x$ with $|f(x)-f(x_0)|\ge \epsilon$. It does not mean that $|x-x_0|\le\delta \Rightarrow |f(x)-f(x_0)|\ge \epsilon$, for example it is false for $x=x_0$.

Yeah, you are right, that error is a fatal one for the above arguments. But nevertheless the solution uses only the definition, after all, it is a problem for high school students. So, take such $\varepsilon>0$ that breaks the definition of continuity and let $I := \left(f(\mathbb{R})\cap (f(x_0)-\varepsilon, f(x_0)+\varepsilon) \right)\setminus f(\mathbb{Q})$. Then, if $f$ is not a constant function (trivially continuous) it follows $I$ is dense at $f(x_0)$. Now, fix some $\delta>0$. There exists $x_1$ at distance no more than $\delta$ from $x_0$ with $|f(x_1)-f(x_0)|\geq \varepsilon$. Depending on the sign of $f(x_1)-f(x_0)$ , we can take $x_2$ near $x_0$ with $y :=f(x_2)\in I$ (by the intermediate value property). Obviously $x_2$ is irrattional. Again, taking $\delta>0$ with $\delta<|x_2-x_0|$ , we can find $x_3$ at distance no more than $\delta$ from $x_0$ with $|f(x_3)-f(x_0)|\geq \varepsilon$. If the sign of $f(x_3)-f(x_0)$ is the same as the sign of $f(x_1)-f(x_0)$ , we can take $x_4$ near $x_0$ with $f(x_4)$ equals the same $y$ , a contradiction. If that signs are different, we can take $x_4$ with $y' := f(x_4)\in I$ and the contradiction is at the next step.

Yes, @dgrozev that's a neat solution. I had some ideas which I mention here, but it is far from being a solution. But possibly we can develop it into a solution by filling gaps and making statements more correct.

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adityaguharoy wrote: In fact an injective Darboux function $f: \mathbb{R} \to \mathbb{R}$ must be monotone. If only we could show this result holds even when $f: \mathbb{R} \setminus \mathbb{Q} \to \mathbb{R}$ then there was a hope of invoking this result. I had the same idea during the contest and so I supposed that $g$ is not monotonic over $\mathbb{R} \setminus \mathbb{Q}$ which means that there exist the numbers $a,b,c \in \mathbb{R} \setminus \mathbb{Q},~a<b<c$ such that $f(b)>\max \{f(a),f(c)\}$ or $f(b)<\min \{f(a),f(c)\}.$ We can assume WLOG that $f(b)>\max \{f(a),f(c)\}.$ It's easy to prove that there are infinitely many numbers $u \neq v$ such that $f(u)=f(v),$ but we don't know what kind of numbers are $u$ and $v$ - and this is the point where I got stuck. I proved the fact that if $g: \mathbb{R} \to \mathbb{R}$ is a Darboux function which is monotone over $\mathbb{R} \setminus \mathbb{Q},$ then $g$ is continuous, but I got $0$ points, which means that this idea might not work at this problem.

My solution during the contest: We will prove that $f$ is monotonic, hence it will be continuous. Suppose that it is not. Then there are $x<y<z$ such that, WLOG, $f(x)<f(y)>f(z).$ Then, for any $\lambda \in I=(f(x),f(y)) \cap (f(z),f(y))$ there are $c_\lambda \in (x,y)$ and $d_\lambda \in (y,z)$ such that $$f(c_\lambda)=f(d_\lambda)=\lambda$$Since $f$ is injective on the irrationals, at least one of these two numbers will be rational, and let $u_\lambda$ be the one. Now define $$g:I \to \mathbb{Q}, \: g(\lambda)=u_\lambda$$We will prove that $g$ is injective. Indeed, let $a,b \in I$ such that $g(a)=g(b) \Rightarrow u_a = u_b \Rightarrow f(u_a)=f(u_b) \Rightarrow a=b.$ Hence there is an injection from $I$ to $\mathbb{Q}.$ Since $\mathbb{Q}$ is countable, there is also an injection from $\mathbb{Q}$ to $\mathbb{N},$ and thus there is an injection from $I$ to $\mathbb{N}.$ But this means that $I,$ a non-empty interval, is countable, which is a contradiction. In conclusion, $f$ is monotonic and so it is continuous.

Question : with the hypothesis can we prove $f^{-1}(\{x\})$ closed for all $x \in \mathbb{R}$ ?

Filipjack wrote: adityaguharoy wrote: In fact an injective Darboux function $f: \mathbb{R} \to \mathbb{R}$ must be monotone. If only we could show this result holds even when $f: \mathbb{R} \setminus \mathbb{Q} \to \mathbb{R}$ then there was a hope of invoking this result. I had the same idea during the contest and so I supposed that $g$ is not monotonic over $\mathbb{R} \setminus \mathbb{Q}$ which means that there exist the numbers $a,b,c \in \mathbb{R} \setminus \mathbb{Q},~a<b<c$ such that $f(b)>\max \{f(a),f(c)\}$ or $f(b)<\min \{f(a),f(c)\}.$ We can assume WLOG that $f(b)>\max \{f(a),f(c)\}.$ It's easy to prove that there are infinitely many numbers $u \neq v$ such that $f(u)=f(v),$ but we don't know what kind of numbers are $u$ and $v$ - and this is the point where I got stuck. I proved the fact that if $g: \mathbb{R} \to \mathbb{R}$ is a Darboux function which is monotone over $\mathbb{R} \setminus \mathbb{Q},$ then $g$ is continuous, but I got $0$ points, which means that this idea might not work at this problem. Yes unfortunately I also could not create a solution from this.