I wan to show that there is no continuous real-valued function of a real variable that sends rationals to irrationals and irrationals to rationals by using only $1)$ and $2)$:
$1)$ We can use the result that if $f$ is continuous then it is sequentally continuous, that is, for every sequence $x_n$ that tends to $c$ the associated sequence $f(x_n)$ tends to $f(c)$ and also the result that if $f$ is continuous then $\lim_{x \to c}f(x)=f(c)$.
$2)$ We can use the result that between every different two numbers on the real line there is an infinite number of irrationals between them and an infinite number of rationals between them.
We should not use anything else, no countability/uncountability arguments, no intermediate value theorem, no that if $A$ is connected then $f(A)$ is connected etc...
Surely $1)$ and $2)$ seem to be too weak to deduce the result but even if they are how to prove it? Can someone explain to me why this can (or cannot) be done by only using $1)$ and $2)$?
If $1)$ and $2)$ are indeed too weak then use also:
$3)$ $\bigcap_{i=1}^{\infty}A_i$ is non-empty if $A_{j+1} \subset A_j$ for every $j \in \mathbb N$ and $A_j$ is non-empty compact set for every $j \in \mathbb N$ and we know that compact means closed and bounded.