Question about the cardinality of a space I've been having conflicting thoughts about the following problem, and I was wondering if anyone could help me out. 

Is is true that the cardinality of every regular separable space does not exceed $2^{\omega}$?

Thank you for your time.
 A: This is false: $\beta\omega$ is a separable compact Hausdorff space of cardinality $2^{2^\omega}>2^\omega$. (Here is a proof that $|\beta\omega|=2^{2^\omega}$.)
In general the best we can say is that if $\langle X,\tau\rangle$ is a Hausdorff space with a dense subset $D$, then $|X|\le 2^{2^{|D|}}$. In particular, is $X$ is separable, $|X|\le2^{2^\omega}$. To see this, for each $x\in X$ let $\mathscr{D}(x)=\{V\cap D:x\in V\in\tau\}$. Since $X$ is Hausdorff, the map $X\to\wp\big(\wp(D)\big):x\mapsto\mathscr{D}(x)$ is injective, so, $|X|\le 2^{2^{|D|}}$. 
A: This is false. I don't know how you're using the assumption of regularity; let me talk about Hausdorff separable spaces $X$ with countable dense subsets $S$. Then a naive argument might run as follows: since $S$ is dense, every element of $X$ is the limit of a sequence with values in $S$, and since $X$ is Hausdorff, limits are unique if they exist. Consequently the cardinality of $X$ is at most the cardinality of the set of sequences with values in $S$, which is $2^{\omega}$.
This argument fails at the very first step: we cannot conclude that every element of $X$ is the limit of a sequence with values in $S$. This is true if $X$ is also first-countable, and then the argument works, but false in general. 
Regular and separable does not imply first-countable. Indeed, any compact Hausdorff space is regular (in fact completely regular by Urysohn's lemma), and an example which is not first-countable and also separable, namely $\beta \mathbb{N}$, was given in the comments. 
