Second-countable implies separable

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

Verbal statement
Every second-countable space is a separable space.

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

Separable space
A topological space is termed separable if it admits a countable dense subset.

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

To prove: There exists a countable dense subset of $$X$$

Proof: We can assume without loss of generality that all the $$B_n$$ are nonempty, because the empty ones can be discarded. Now, for each $$B_n$$, pick any element $$x_n \in B_n$$. Let $$D$$ be the set of these $$x_n$$s. $$D$$ is clearly countable (because the indexing set for its elements is countable). We claim that $$D$$ is dense in $$X$$.

To see this, let $$U$$ be any nonempty open subset of $$X$$. Then, $$U$$ contains some $$B_n$$, and hence, $$x_n \in U$$. But by construction, $$x_n \in D$$, so $$D$$ intersects $$U$$, proving that $$D$$ is dense.

Textbook references

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