Let $\pi: \mathbb{R}\to \mathbb{R}/\mathbb{Q}, x \mapsto x+\mathbb{Q}$ be the canonical projection map. Set $$\{1,\sqrt{2}\}/\mathbb{Q}:= \pi(\{1,\sqrt{2}\}) = \{1+\mathbb{Q},\sqrt{2}+\mathbb{Q}\}\subseteq \mathbb{R}/\mathbb{Q}$$ with subspace topology.
Define the map $f:\{1+\mathbb{Q},\sqrt{2}+\mathbb{Q}\}\to \mathbb{R}$ by $1+\mathbb{Q} \mapsto 1$ and $\sqrt{2}+\mathbb{Q}\mapsto \sqrt{2}$. Then $f$ is continuous by the universal property of quotient topology because the inclusion $$\iota: \{1,\sqrt{2}\}+ \mathbb{Q} \subseteq \mathbb{R} \to \mathbb{R},\ x \mapsto x$$ is continuous, $$\pi|_{\{1,\sqrt{2}\}+ \mathbb{Q}}: \{1,\sqrt{2}\}+ \mathbb{Q} \to \{1,\sqrt{2}\}/\mathbb{Q}$$ is a quotient map, because it is open, surjective and continuous, and $f\circ \pi|_{\{1,\sqrt{2}\}+ \mathbb{Q}} = \iota$.
Where is this "proof" wrong?
My "proof" cannot be true, since $\{1,\sqrt{2}\}/\mathbb{Q}\subseteq\mathbb{R}/\mathbb{Q}$ and $\mathbb{R}/\mathbb{Q}$ has the trivial topology, but it should be discrete, if $f$ were continuous.
I know that is a silly question, but I cannot see at the moment where I am wrong?
The quotient topology on $\Bbb{R}/\Bbb{Q}$ induced by $\pi$ is the indiscrete topology. Its only open sets are $\Bbb{R}/\Bbb{Q}$ and $\emptyset$. If you take $B=\{1+\Bbb{Q},\sqrt{2}+\Bbb{Q}\}$, as a subspace of $\Bbb{R}/\Bbb{Q}$, it also has the indiscrete topology where its only open sets are $B$ and $\emptyset$. The map $f:B\to\Bbb{R}$ such that $f(1+\Bbb{Q})=1$ and $f(\sqrt{2}+\Bbb{Q}) = \sqrt{2}$ is discontinuous because $\{1,\sqrt{2}\}$ have the discrete topology as a subset of $\Bbb{R}$. So $f^{-1}(\{1\})=\{1+\Bbb{Q}\}$ and $f^{-1}(\{\sqrt{2}\})=\{\sqrt{2}+\Bbb{Q}\}$ are not open subsets of $B$ - which they should be.