Second-countable implies Lindelof

Property-theoretic statement
The property of topological spaces of being a second-countable space implies, or is stronger than, the property of being a Lindelof space.

Verbal statement
Any second-countable space is Lindelof.

Second-countable space
A topological space is termed second-countable if it admits a countable basis.

Lindelof space
A topological space is termed Lindelof if every open cover of the space has a countable subcover.

Proof
Given: A second-countable space $$X$$ with countable basis $$\{ B_n \}$$

To prove: If $$\{ U_i \}_{i \in I}$$ form an open cover of $$X$$, there exists a countable subcover of $$X$$ among the $$\{ U_i \}$$s

Proof: For each basis element $$B_n$$, let $$V_n$$ be any $$U_i$$ containing $$B_n$$, if such a $$U_i$$ exists, otherwise, pick nothing. This gives a (at most) countable subcollection $$\{ V_n \}$$ of the collection $$\{ U_i \}$$. We want to show that this subcollection covers $$X$$.

Suppose there exists $$x \in X$$, that does not belong to any $$V_n$$. Then, since the entire collection of $$U_j$$ cover $$X$$, we can find some $$j \in I$$ such that $$x \in U_j$$. Further, there exists some $$n$$ such that $$x \in B_n \subset U_j$$. Now, since $$B_n$$ is contained in at least one of the $$U_i$$s, there should exist an element $$V_n$$. But such an element would also contain $$x$$, contradicting the claim that $$x$$ does not belong to any $$V_n$$.

Textbook references

 * , Page 191, Theorem 30.3(a), Chapter 4, Section 30