Countable Chain Condition for separable spaces? 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.
 A: The Suslin line is a topological space which has the CCC property but is not separable.
However, proving that the Suslin line exists cannot be done within $ZFC$. Why?
Assuming $V=L$ (The axiom of constructibility) implies certain combinatorial properties from which we can construct a Suslin line, however assuming a different axiom $MA$ (Martin's axiom) we can prove that no Suslin line exists.

The result of this is that we cannot prove from ZFC alone that every CCC space is separable.
Added: (To make this answer complete, I'll add the right answer given by Henno Brandsma in the comments)
We cannot prove in ZFC that CCC spaces are separable because $\{0,1\}^X$ has CCC for any $X$, but is only separable for $|X|\le\frak c$. In particular, taking $X=P(\mathbb R)$ gives us a CCC space which is not separable.
A: It is explained in this MathOverflow question that there are many spaces that are ccc but not separable, including any products of ccc separable spaces with sufficiently many nontrivial factors. In particular, there is no ZFC independence on this question: we can prove that there are ccc spaces that are not separable.
Other examples include the numerous instances of ccc forcing notions that have no countable dense set, which are quite commonly considered in forcing arguments.
A: 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.

A: My understanding is that in the class of all topological spaces, there exist ccc spaces which are not separable (and there are examples (*) of these).  But, if we restrict our attention to spaces with the order topology, the question is undecidable in ZFC (See the Suslin Hypothesis).  
Assuming Martin's Axiom (+$\neg$CH), SH is true (for spaces with the order topology, ccc implies separable).  It can be shown that a product of non-separable ccc spaces is not ccc, yet MA implies products of ccc spaces are ccc.
So in particular, the spaces in the examples * must have topologies which are not equivalent with any order topology on the underlying set.
