For any irrational number $x$, let $f(x)$ be the real number arising from the integer part of $x$, together with every other digit of the rest of the expansion of $x$.

This is surjective, since one may interleave the digits of any real $y$ with any nonrepeating pattern, and thereby find an irrational $x$ with $f(x)=y$. This is continuous, since in order to determine finitely many digits of $f(x)$, it suffices to specify finitely many digits of if $x$.

Oops, it's not actually continuousx_n\to x$, because of non-unique representationthen $f(x_n)\to f(x)$.

For any irrational number $x$, let $f(x)$ be the real number arising from the integer part of $x$, together with every other digit of the rest of the expansion of $x$.

This is surjective, since one may interleave the digits of any real $y$ with any nonrepeating pattern, and thereby find an $x$ with $f(x)=y$. This is continuous, since in order to determine finitely many digits of $f(x)$, it suffices to specify finitely many digits of $x$.

Oops, it's not actually continuous, because of non-unique representation.

For any irrational number $x$, let $f(x)$ be the real number arising from the integer part of $x$, together with every other digit of the rest of the expansion of $x$.

This is surjective, since one may interleave the digits of any real $y$ with any nonrepeating pattern, and thereby find an $x$ with $f(x)=y$. This is continuous, since in order to determine finitely many digits of $f(x)$, it suffices to specify finitely many digits of $x$.