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$(X,d)$ 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.

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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.

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    I tried to do the proof but I am stuck in showing the sets are nowhere dense;Please help. – 2017-01-09
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    each of the sets are closed since they are inverse images of closed sets but how to show their interior is empty – 2017-01-09
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    So if there are no points with both irrational cooridantes in the image of $f$, then $X=\bigcup H_q \cup \bigcup K_q$. But $K_q$ and $H_q$ are first category sets, so sum of countable many of such sets is also first category in complete metrics spaces. But $X$ is not first category set, therefore there must be point with irrational cooridantes. Right? – 2017-01-09
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    @learnmore: Consider an $H_q$, for instance, and let $U$ be its interior. If $U\ne\varnothing$, then $f[U]$ is a non-empty subset of the straight line $\{q\}\times\Bbb R$, which is impossible. This means that $H_q$ is a closed set with empty interior and hence is nowhere dense. – 2017-01-09
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    @SekstusEmpiryk: Yes, that’s right. – 2017-01-09
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    yes i got that;thank you so much @BrianM.Scott – 2017-01-09
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    @learnmore: You’re welcome. – 2017-01-09
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    @BrianM.Scott; I am studying metric space and topology;you being an expert in this field will you please help me how to grasp this subject or how did you expertise in this topic,any books or references – 2017-01-09