CCC means countable chain condition;
A cover $\cal A$ of a set $E$ is separating if for each $p\in E$, $\bigcap \{A: A \in \mathcal{A}, p\in A\}=\{p\}.$ The point separating weight of $X$, denoted $psw(X)$, is the smallest infinite cardinal $\kappa$ such that $X$ has a separating open cover $\cal V$ with ord$(p, \cal V)\le \kappa$ for each $p\in X$.
Let $X$ be a CCC space and has countable point separating weight. Is the cardinality of $X$ always at most $2^\omega$?
Thanks for any help:)
$\newcommand{\supp}{\operatorname{supp}}$The cardinality of the space can by arbitrarily large.
Let $\kappa$ be any infinite cardinal. Let $\varphi:[\kappa]^{<\omega}\to\kappa$ be any injection with the property that $\varphi(F)\notin F$ for each $F\in[\kappa]^{<\omega}$. For each $\xi<\kappa$ let $X_\xi$ be a copy of $[0,1]$, and let $X=\prod_{\xi<\kappa}X_\xi$, with the product topology. For $F\in[\kappa]^{<\omega}$ let
$$X_F=\left\{x\in X:x_{\varphi(F)}=1\text{ and }\forall\xi\in F\Big(x_\xi\in(0,1)\Big)\text{ and }\forall\xi\in\kappa\setminus\big(F\cup\{\varphi(F)\}\big)(x_\xi=0)\right\}\;.$$
Finally, let $$Y=\bigcup_{F\in[\kappa]^{<\omega}}X_F$$
with the subspace topology inherited from $X$, and for each $x\in Y$ let $\supp(x)$ be the unique $F\in[\kappa]^{<\omega}$ such that $x\in X_F$.
Let $\mathscr{I}=\{(p,q):p<q\text{ and }p,q\in\Bbb Q\cap(0,1)\}$, and let $\mathscr{J}=\{(p,1]:p\in\Bbb Q\cap(0,1)\}.$ For each $F\in[\kappa]^{<\omega}$, function $I:F\to\mathscr{I}$, and $J\in\mathscr{J}$ let
$$B(F,I,J)=\left\{x\in Y:\forall\xi\in F\Big(x_\xi\in I(\xi)\text{ and }x_{\varphi(F)}\in J\Big)\right\}\;,$$
and let $$\mathscr{B}=\left\{B(F,I,J):F\in[\kappa]^{<\omega}\text{ and }I:F\to\mathscr{I}\text{ and }J\in\mathscr{J}\right\}\;;$$ clearly $\mathscr{B}$ is an open cover of $Y$. Suppose that $x\in B(F,I,J)\in\mathscr{B}$. Then $x_\xi\ne 0$ for each $\xi\in F\cup\{\varphi(F)\}$, so $F\cup\{\varphi(F)\}\subseteq\supp(x)$, and it follows immediately that $\operatorname{ord}(x,\mathscr{B})=\omega$. It’s also clear that for each $x\in Y$ we have $\bigcap\{B\in\mathscr{B}:x\in B\}=\{x\}$, so $\operatorname{psw}(Y)=\omega$.
Finally, $X$ is a product of separable spaces, so $c(X)=\omega$, and $Y$ is dense in $X$, so $c(Y)=\omega$ as well.