Does separability imply the Lindelöf property? Does separability imply a sort of Lindelöf property?
Since I can't prove this fact I'm beginning to think that my conjecture is false.
Intuitively, $\mathbb{R}$ has a countable subset $\mathbb{Q}$ which is used to form a countable basis for $\mathbb{R}$ with the usual topology and prove the Lindeöf property.
 A: Counterexamples in Topology lists several spaces which are separable but not Lindelöf.  You can generate a list at Spacebook.
A: A nice example that is useful to know is the Mrówka space $\Psi$, which among other things is Tikhonov, separable, pseudocompact, not countably compact, and not Lindelöf. I also described it briefly in this answer, where I gave an example of a specific open cover with no countable subcover. This answer contains a more complete description of the construction of the space, and this one has another way to get the almost disjoint sets that are needed.
A: Every seperable meta-Lindelöf (or meta-compact) space is Lindelöf.
We say a space is meta-Lindelöf (meta-compact)  if every open cover has a point countable(finite) open refinement. That is, given any open cover of the topological space, there is a refinement which is again an open cover with the property that every point is contained only in countably(finitely) many sets of the refining cover.
Proof:
Given a space $X$, and an open cover $\mathfrak U$ of $X$, by the meta-Lindelöf property get a point countable open refinement $\mathfrak V$ of $\mathfrak U$. So, for every $x\in X$ the set $\mathfrak V_x=\{V\in\mathfrak V:x\in V\}$ is countable. If $D$ is a countable dense subset of $X$, observe that $\mathfrak V=\bigcup_{x\in D} \mathfrak V_x$, so $\mathfrak V$ is countable. Now, since $\mathfrak V$ is an open refinement of $\mathfrak U$, for each $V_n\in \mathfrak V$ exists $U_n\in\mathfrak U$ for each $n\in\mathbb N$, so the cover $\mathfrak W=\{U_n:n\in\mathbb N\}$ is a countable subcover of $\mathfrak U$. So $X$ is Lindelöf. 
