Suppose $(X,\tau)$ is a topological space and $A\subset X$,suppose $\{G_\alpha\}$ is an open cover of $A$ i.e. $A\subset \cup G_\alpha$,does there exist a countable subcover $\{G_{\alpha_n}\}$ of $A$.I know that if $X$ is a metric space then $X$ is Lindelof will imply $(A,d_A)$ is Lindelof then $\{G_\alpha\cap A\}$is an open cover of $A$.Then it has a countable subcover $\{G_{\alpha_n}\cap A\}$,so that $A=\cup G_{\alpha_n}\cap A \subset \cup G_{\alpha_n}$,So $A$ has a countable subcover.But is it true for arbitrary topological space because we are using metric property on the line marked with bold letters.
Addendum I think I was confused with a trivial thing.If $Y\subset X$ is Lindelof then Every $X$-open cover of $Y$ has a countable $X$-open subcover and conversely.This fact is trivial in fact but I am a beginner,so I could not see it at first.Actually my question was whether this think is true.
This is false in general and spaces for which it holds are called hereditarily Lindelöf.
A counterexample is given by the ordinal $\omega_1+1$ in the order topology, this space is compact hence Lindelöf, but the subspace $\omega_1$ is not: the cover of initial segments $ \{[0,\xi)\mid \xi<\omega_1\}$ has no countable subcover, since every countable union of countable ordinals is a countable ordinal, so bounded in $\omega_1$.