Can a vector space with infinite basis be represented as a countable union of proper subspaces?

Question

I am familiar with the case where the basis is not infinite (it becomes the case that it CANNOT be represented as such). However, many sources suggest that this is not the case for an infinite basis. How could one go about proving this?

Answer 1

Suppose that $V$ is infinite dimensional and fix a basis $B$. Then, by assumption, we have $|B| \geq \aleph_0$, and thus there exists a countable subset $S \subset B$. For each $s \in S$, consider $V_s = \langle B \setminus \{s\} \rangle$. Then $V = \bigcup_{s \in S}V_s$.