$f:X\rightarrow\mathbb{R}^2$ continuous from complete metric space, must have in image point with irrational cooridantes $(X,d)$ is complete metric space. We have $f:X\rightarrow\mathbb{R}^2$, which is continuous and, for any open set in $X$, its image is not included in straight line in $\mathbb{R}^2$. Prove that there exist $x$ such that $f(x)=(x_1,x_2)$ where $x_1,x_2\in\mathbb{R}\setminus\mathbb{Q}$
I need advice. I suspect Baire theorem must be used there.

HINT: Yes, the Baire category theorem can be used. For each $q\in\Bbb Q$ let $H_q=f^{-1}[\{q\}\times\Bbb R]$ and $K_q=f^{-1}[\Bbb R\times\{q\}]$. Show that each of the sets $H_q$ and $K_q$ is closed and nowhere dense in $X$, and note that there are only countably many such sets.
