Is the set of irrationals separable as a subspace of the real line? I am trying to find an example of a separable Hausdorff space which has a non-separable subspace.  This led me to ask the question in the title: is the set of irrationals, regarded as a subspace of the real line, separable or  non-separable?
A space is separable if it contains a countable dense subset.  A subset $A$ of a space $X$ is dense in $X$ if $\bar{A}=X$.
It's easy to come up with a dense set and a countable set in $\mathbb{R}\setminus\mathbb{Q}$, since (trivially) $\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\setminus\mathbb{Q}$ in the subspace topology, and as a countable set we can take something like $\{k\pi \mid k \in \mathbb{Z}\}$.  But is there a subset that is both dense AND countable?  And of course, how do we prove the result?
 A: $\Bbb R$ is second countable (i.e., has a countable base), so it’s hereditarily separable. Specifically, let $\mathscr{B}$ be the set of all open intervals with rational endpoints; $\Bbb Q$ is countable, so $\mathscr{B}$ is countable. Enumerate $\mathscr{B}=\{B_n:n\in\Bbb N\}$, and for $n\in\Bbb N$ let $x_n$ be any irrational number in $B_n$. Then $\{x_n:n\in\Bbb N\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$. (Clearly the same trick works for any subset of $\Bbb R$, not just the irrationals.)
For an explicit example of such a set, let $\alpha$ be any irrational; then $\{p+\alpha:p\in\Bbb Q\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$.
There are many separable Hausdorff spaces with non-separable subspaces. Two are mentioned in this answer to an earlier question. The first is compact; the second is Tikhonov and pseudocompact. Both are therefore quite nice spaces. Both are a bit complicated, however. A simpler example is the Sorgenfrey plane $\Bbb S$: $\Bbb Q\times\Bbb Q$ is a countable dense subset of $\Bbb S$, and $\{\langle x,-x\rangle:x\in\Bbb R\}$ is an uncountable discrete subset of $\Bbb S$ (which is obviously not separable as a subspace of $\Bbb S$).
A: Every subspace of a separable metric space is separable.
For the irrationals, take the irrational algebraic numbers, those are dense in $\mathbb R$ and therefore in the irrationals too. As Jacob remarks below, if $\alpha$ is irrational, then $\{\alpha+q\mid q\in\mathbb Q\}$ is also dense.
Generally speaking, the irrationals are homeomorphic to the Baire space, the set of sequences of natural numbers, equipped with the product topology $\mathbb N^\mathbb N$. 
This has a countable dense subset: eventually constant sequences.
