Let $x \sim y$ if and only if $x-y\in \mathbb{Q}$.Show that the quotient topology on $\mathbb{R}/\sim$ is the indiscrete topology. Consider $\mathbb{R}$ with standard (Euclidean) topology.
Let
$x \sim y$ if and only if $x-y\in \mathbb{Q}$.
Show that $\mathbb{R} /\sim$ is uncountable but that the quotient topology on $\mathbb{R}/\sim$ is the indiscrete topology.
To show that $\mathbb{R} /\sim = \{[x] : x\in \mathbb{R}\}$ is uncountable, suppose that it is countable. 
Then the equivalence classes disjointly partitions $\mathbb{R}$. 
Since $[x] = \{y\in \mathbb{R} : y\sim x\} = \{x+q: q\in \mathbb{Q}\}$ is countable, therefore $\mathbb{R}$ is the countable union of countable set. Since countable union of countable set is countable, $\mathbb{R}$ is then countable and this is a contradiction. Hence $\mathbb{R} /\sim = \{[x] : x\in \mathbb{R}\}$ is uncountable.
To show that the quotient topology on $\mathbb{R} /\sim$ is the indiscrete topology, let's prove that for any nontrivial strict open subset $U$ of $\mathbb{R} /\sim$, we have $$\bigcup_{[x]\in U} [x]$$ being closed in $\mathbb{R}$.
When $U$ is finite, it is closed in $\mathbb{R}$ since each $[x]$ is closed in $\mathbb{R}$ and finite union of closed sets is closed. 
However, I am not sure how to complete the proof when $U$ is any arbitrary subset.

A different attempt:


*

*Recall that a set in $\mathbb{R} / \sim$ is open iff its preimage under the projection is open.

*Show that every open set in $\mathbb{R}$ contains a representative of every residue class.

