I have been looking at Lindelof spaces and searched for some properties.
So far I know that the product of Lindelof spaces needs not neccessarily be a Lindelof space(Sorgenfey plane), neither needs the product of a Lindelof space and a metrizable space be Lindelof.
Further the Lindelof property implies the existence of a countable basis only if the space is metrizable.
Let $X = \mathbb{S} \times\mathbb{R}$ where $\mathbb{S}$ is the Sorgenfey line, i.e. the topological space generated by the half open intervals $\left] a , b \right]$ and $\mathbb{R}$ induced with the usual euclidean topology. I think that X is not metrizable and does not have a countable basis but I still think it might be Lindelof, although I cannot really prove it rigorously.
Definition: A topological space is called Lindelof, if for every open cover there exists a countable subcover.
Thanks in advance!
Yes, it is Lindelöf, because $\mathbb{R}$ in the usual topology is $\sigma$-compact, and when $X$ is Lindelöf and $C$ is compact, then $X \times C$ is Lindelöf. So $\mathbb{S} \times \mathbb{R}$ is a countable union of Lindelöf subspaces, and hence Lindelöf too.
A Michael-line construction $\mathbb{R}_B$ based on a Bernstein set $B$ has the property that $\mathbb{R}_B \times B$ is not normal (so not Lindelöf as well), while $\mathbb{R}_B$ is hereditarily Lindelöf and $B$ is separable metrisable (so second countable). This shows we need "extreme Lindelöfness" (e.g. in the form of being $\sigma$-compact) to preserve Lindelöfness in a product..
The product is clearly not metrisable nor second countable as its subspace $\mathbb{S}$ is neither and both properties are hereditary.