Constructing a separable space that is not hereditarily separable. The construction that I had in mind was:
Let $X$ be an uncountable space. We'll assume it has just one countable dense subset $E$. Let points $p_{1},p_{2}\in X\setminus E$ such that every open set containing either of the two points contains some or all points $a_{1},a_{2}\dots a_{n}\in E$. Then the space $X-\{ a_{1},a_{2}\dots a_{n}\}$ won't have a countable dense subset, as $E-\{ a_{1},a_{2}\dots a_{n}\}$ won't have $p_{1},p_{2}$ in its closure. 
Am I making too many assumptions?
Thanks for your time!
 A: This is not so much a construction, as a vague idea, maybe. I suppose you want a separable but not hereditarily separable space? 
A simple example $X$ is $\mathbb{R}$, in the included point topology with respect to $0$. So $O$ is open iff it is empty or it contains $0$. Then $\{0\}$ is dense, so $X$ is separable, but $X\setminus\{0\}$ is discrete in itself, so not separable. But $X$ is not even $T_1$.
Nicer examples: the $S \times S$, where $S$ is the Sorgenfrey line: this is separable, but has the set $D = \{(x, -x): x \in \mathbb{R} \}$ as a closed and discrete subspace. Or any uncountable product like $\mathbb{R}^{[0,1]}$, which is separable, but the set of all points with exactly one value equal to $1$ and all others to $0$, is again discrete as a subspace (and uncountable), so we have another non-separable subspace. 
A: This may not warrant an answer, but here are couple observations.


*

*You will have to be careful in your assumption of "just one countable dense subset" since adding any point to a countable dense set will result in another.  (Of course, a countable discrete space really does have only one countable dense subset, but a separable, not hereditarily separable space must be uncountable.)

*It is not clear why $E_0 = ( E \setminus \{ a_1 , \ldots , a_n \} ) \cup \{ p_1 , p_2 \}$ couldn't be a countable dense subset of $X \setminus \{ a_1 , \ldots , a_n \}$.

