= Solution
Suppose $(x_k)_{k\geq1}$ were a countable <Hamel basis>. Put $E_n=\operatorname{span}(x_1,\ldots,x_n)$. Each $E_n$ is a finite-dimensional, hence closed, <linear subspace> of $X$. It is proper because $X$ is infinite dimensional, and therefore a <nowhere dense set>. A <Hamel basis> means that every vector is a finite <linear combination>, so $X=\bigcup_nE_n$. This contradicts the <Baire category theorem> for the <Banach space> $X$. Hence \b[an infinite-dimensional Banach space has no countable Hamel basis]. The same proof excludes any countable algebraic spanning set.
Back to article page
