TITLE: Countable Chain Condition for separable spaces?
QUESTION [5 upvotes]: I'm trying to find a proof behind a small proposition. Recall that a topological space satisfies the countable chain condition if each disjoint collection of open sets is countable.

Why is it the case that every separable spaces satisfies the CCC, but the converse is not true?

Thanks.

REPLY [2 votes]: I've been trying to come up with a simpler example to answer your question, although the answes of course complete and great, I still fancied a simpler example. So here it is:

Every separable space is CCC. The converse is not true.

For example:

Let $X$ be a space with $|X|= \aleph_1$, let $\tau_X= \lbrace U: X\setminus U \text{ is countable } \rbrace$. This space is CCC, but not separable.
Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$.
$X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open.