Quotient space of the reals by the rationals

It's easy to see that the space is connected. Since every non-empty open set in $\Bbb R$ contains rationals, there is a point which meets every open set of $\Bbb R/{\sim}$. So if $U\cap V=\varnothing$ are open sets, at least one has to be empty.

Next, note that not being closed does not mean being open. The irrational numbers are not closed since they are still dense in the space, but they are not open either because the rational "point" is not closed (how could it be closed? It's dense!).

I'll add my own answer, elaborating on my comments to the original post and to Rob Arthan's (now deleted) answer, just to clear things up.

A subset of $\mathbb{R}/{\sim}$ is open if and only if its preimage under the map $\mathbb{R} \to \mathbb{R}/{\sim}$ is open. Since every nonempty open set in $\mathbb{R}$ contains a rational number, the open sets of $\mathbb{R}/{\sim}$ are in bijection with $\{U\subseteq \mathbb{R}\mid U\text{ is open, and }\mathbb{Q}\subset U\}\cup \{\emptyset\}$.

How can we find open sets in $\mathbb{R}$ that contain $\mathbb{Q}$? Well, such a set is exactly the complement of a closed set which is disjoint from $\mathbb{Q}$, i.e. a closed set of irrationals. So we can take $\mathbb{R}$ and remove a single irrational, or finitely many irrationals, or a sequence of irrationals limiting to some irrational, etc. etc.

The somewhat unintuitive thing is that there are some very large (in the sense of Lebesgue measure) closed sets of irrationals. For example, fix $\epsilon>0$, enumerate the rationals as $\{q_n\mid n\in\mathbb{N}\}$, and let $O_n$ be an open interval of length $\epsilon \cdot 2^{-n}$ containing $q_n$. Then let $U = \bigcup_{n\in\mathbb{N}} O_n$. Then $\lambda(U) \leq 2\epsilon$, where $\lambda$ is the Lebesgue measure.

To see that the image of the set $I$ of irrationals in $\mathbb{R}/{\sim}$ is not compact, let $U$ be the set defined above for $\epsilon <\frac{1}{2}$ (so $U$ doesn't contain any interval $(n,n+1)$). Then for $n\in \mathbb{Z}$, let $U_n = U \cup (n,n+1)$. Now each $U_n$ is open, and $I\subseteq \bigcup_{n\in\mathbb{Z}} U_n$, but this open cover has no finite subcover. All the $U_n$ contain $\mathbb{Q}$ so the same is true of their images in $\mathbb{R}/{\sim}$.