separability of a space When I have to show that some space $A$ IS NOT separable, does it always work if I find uncountable subset $B\subset A$, $|B|=2^{\aleph_0}$ and set C of disjoint  open balls, $C=\{L(x,r): x\in B\}$.  
If $A$ IS separable, then $\exists D\subset A$ such that $|D|\leq \aleph_0$ and $(\forall x\in A)(\forall r >0)(\exists y\in D) y\in L(x,r)$.  
So, if I want A to be separable, then $D$ has to have non-empty intersection with all elements from $C$, so it would be $|D|=|C|=|B|=2^{\aleph_0}$, which means that D does not satisfy condition $|D|\leq \aleph_0$ and $A$ is not separable.
 A: Yes, you are absolutely right.  Any dense set must intersect every nonempty open set.  So if you can find uncountably many nonempty open sets that are pairwise disjoint, then any dense set must be uncountable, and the space cannot be separable.
A: A topological space $X$ is said to have the countable chain condition (or to be ccc) if every family of pairwise disjoint nonempty open subsets of $X$ is countable.
What you are essentially demonstrating is that every separable space is ccc: Letting $A$ be a countable dense set, then if $\{ U_i : i \in I \}$ were uncountable family of pairwise disjoint nonempty open sets, then we may pick $x_i \in U_i \cap A$ for each $i$.  But the $x_i$ must be pairwise distinct (since the $U_i$ were pairwise disjoint), which contradicts that we only had countably many of them to begin with!
In metric spaces, it turns out that being ccc is equivalent to being separable.
However, for general topological spaces ccc-ness is not equivalent to separability, even for "nice" spaces.  For example, letting $\mathfrak{c} = 2^{\aleph_0}$ denote the cardinality of the continuum, one can show that $$X = \{ 0 , 1 \}^{\Large 2^{\mathfrak{c}}}$$ is ccc, but it is not separable.  (We give $\{ 0 , 1 \}$ the discrete topology, and then give $X$ the product topology on $2^{\mathfrak{c}}$ copies of $\{ 0 , 1 \}$.  Note that $X$ is compact and Hausdorff, and thus normal, so by many measures it's a quite nice space.)
