Is every dense subspace of a separable space separable? If $X$ is a separable topological space, and $V$ is some dense subspace of it, is $V$ necessarily separable?
 A: No. Let $X=\omega_1+1$, where $\omega_1$ is the first uncountable ordinal. Let $U\subseteq X$ be open iff $U$ is empty or if it contains some uncountable tail, i.e. if $\{\alpha;\alpha>\beta\}\subseteq U$ for some $\beta<\omega_1$. One easily checks that this is a topology. 
Now observe that the point $\omega_1$ is in every nonempty open set and therefore $X$ is the closure of this single point and thus separable. In fact, it's not hard to see that a subspace $Y$ is dense in $X$ iff $Y$ is unbounded. So consider the subspace $Y=\omega_1\subseteq X$. By our observation $Y$ is dense in $X$ but is not itself separable since no countable set can be unbounded in $Y$.
A: A classical example of this: let $X = \prod_{i \in I} [0,1]_i$, where $I$ has size $|\mathbb{R}|$ and all $[0,1]_i$ are copies of $[0,1]$. Then $X$ is a compact separable set (Hewitt-Marczewski-Pondizcery theorem), but the sigma product $Y = \Sigma_{i \in I} [0,1]$ (all $(x_i) \in X$ with such that all but countably many coordinates are non-zero) is dense in $X$, but not separable, as every countable set in $Y$ only depends on countably many coordinates, and so cannot be dense. $Y$ is also countably compact (and not compact, of course), for similar reasons.
