Let $F\subset\mathbb{R}$ be a closed nowhere dense set. One must show there exists $(a,b)\in S^1$ for which $b\neq qa+c$, for all $q\in\mathbb{Q},c\in F$.
It's my second question concerning Baire category theorem - I already know what the crux is -- to find a family of closed nowhere dense sets, preferably (in this example) connected with $q,c$ and thus with the set $F$. But no such family comes to my mind. I suppose though, it should comprise some subsets of $F$. I wanted to thoroughly think about it, so what I need is just a little hint.
Suppose , that assertion is not true. This means that for every $(a,b)\in\{(x,y)\in\mathbb{R}^2 :x^2 +y^2 =1\}=S^1$ there exists a rational number $q\in\mathbb{Q}$ and $c\in F$ such that $b=qa+c$. For a given rational number $q\in \mathbb{Q}$ denote by $A_q =\{(u,v)\in S^1 :v-qu\in F\} .$ Since $S^1 = \bigcup_{q\in\mathbb{Q}} A_q$ and $S^1$ is complete metric space ( with induced from $\mathbb{R}^2$ Euclidean metric ) thus from Baire'a Theorem there exists $q_0 \in \mathbb{Q}$ such that the set $A_{q_0}$ is dense in some ball on the circle $S^1 .$ But this means that the set $\overline{A_{q_0}}$ contain some open arc of this circle. Let us now define a map $f:S^1 \rightarrow \mathbb{R}$ by $f(x,y) =y-q_0 \cdot x .$ Since $f$ is continuous $f(\overline{A_{q_0}} ) \subset \overline{f(A_{q_0} )} \subset \overline{F} =F$ but the last shows that the set $F$ contains a connectet subset with two or more elements, so $F$ must contains an open interval. Contradiction.