Does f ( [ 0 , 1 ] ) must always contain an irrational number?

$[0,1]$ is uncountable, and so because $f$ is injective, so will $f([0,1])$. If $f([0,1])$ did not contain an irrational number, then it would lie in $\mathbb{Q}$, and so be countable. But that would give us a contradiction.

Because $[0,1]\setminus \mathbb{Q}$ has the same cardinality as $[0,1]$, there is in fact a bijection from $[0,1]$ onto $[0,1]\setminus \mathbb{Q}$.

$f(x)=\{\begin{array}{ll}\{\pi +x\}& x\text{ is of the form }n\pi +q\text{ for }n\text{ non-negative integer and }q\text{ rational}\\ x& \text{ otherwise}\end{array}$
where $\{x\}$ is the fractional part of $x$, though strictly speaking this is a bijection $[0,1)\to [0,1)\mathrm{\setminus }\mathbb{Q}$