Separable Space satisfies Countable Chain Condition

Theorem
Let $T \mathop = \struct {S, \tau}$ be a topological space.

If $T$ is separable, then $T$ satisfies the countable chain condition.

Proof
Because $T$ is separable there exists a subset $\set {y_n : n\mathop\in \mathbb{N}}$ of $S$ which is everywhere dense in $S$.

Now consider a indexed family $ \family {U_j}_{j \mathop \in J}$ of non-empty open sets such that $U_i\cap U_j\mathop =\emptyset$ for all $i,j\mathop \in J$, $i\mathop \neq j$.

Using Equivalence of Definitions of Everywhere Dense this implies that for every $j\mathop \in J$ there has to exist $n_j\mathop \in\mathbb{N}$ such that $y_{n_j}\mathop \in U_j$.

This gives rise to a well-defined mapping $f:J\to\mathbb N$ via $\map f j:=n_j$.

In particular $f$ is injective: if there existed $i,j\in J$, $i\neq j$ such that $n_i\mathop =n_j$, then $y_{n_i}\mathop \in U_i\cap U_j$, but the latter is the empty set by assumption, a contradiction.

Therefore $J$ is countable by definition.

This concludes the proof.