Why does "separable" imply the "countable chain condition"? Why does "separable" imply the "countable chain condition"?
Thanks for any help.
 A: The reason is that separability implies that there is a countable subset $D$ that for every non-empty open set $U$, $D\cap U\neq\varnothing$.
If $X$ does not have CCC then there is an uncountable family $\{U_i\mid i\in I\}$ of pairwise disjoint open sets. Any countable set can intersect only countably many of those, but never all of them, therefore $X$ is not separable.
Alternatively, you can argue directly: If $\mathcal{U}$ is a family of pairwise disjoint open sets, each $U \in \mathcal{U}$ must contain a point $d \in D$ by density, so there's a surjection from some subset of $D$ onto $\mathcal{U}$, hence $\mathcal U$ is countable.
The other direction is not true since there are CCC spaces which are not separable.
An example of a non-separable but CCC space would be a sufficiently high power $\{0,1\}^\kappa$ ($\kappa \gt \mathfrak{c}$ is enough): indeed, an arbitrary product $\prod_{i \in I} X_i$ of topological spaces is CCC if all finite products $\prod_{j \in J} X_j$ with $J \subset I$, $\# J \lt \infty$ are CCC.
A: Let $D$ be a dense subset of $X$ and let $\{ U_i : i \in I \}$ be a pairwise disjoint family of non-empty open sets indexed by $I$. 
Define a map $f: I \rightarrow D$ by picking $f(i) \in U_i \cap D$, which can be done as each $U_i$ is non-empty and open and $D$ is dense. For a countable $D$ we can choose the one with minimal index in some fixed enumeration of $D$, for definiteness.
The function $f$ is 1-1, because if $i \neq j$ then $f(i) \in U_i$ and $f(j) \in U_j$ but as $U_i \cap U_j = \emptyset$, $f(i) \neq f(j)$. 
Hence we have an injection from $I$ into $D$ and so $|I| \le |D|$ as cardinal numbers.
If $X$ is separable we can fix some countable dense subset $D$ and this then shows that
all pairwise disjoint families of non-empty open sets are at most countable, or $X$ is ccc.
