Basis of $\R^\infty=\R^\N$ must be uncountable

Recall that a subset $M$ of a vector space $X$ is a basis if and only if every element in $X$ can be written as a linear combinations of finitely many elements in $M$.

The title of this post has a simple proof: $\{a_\alpha\in \R^\infty\}_{\alpha>0}$, where $a_\alpha = (1^\alpha,2^\alpha,3^\alpha,\dots)$, is an uncountable linearly independent subset, so every basis has to be uncountable.

I just want to point out knowledge in functional analysis quickly gives an intuitive answer.

Suppose $\R^\infty$ has $M$ as a countable basis, then consider its vector subspace \[\ell^\infty:=\{(x_1,x_2,\dots)\in \R^\infty: \{x_n\} \text{ bounded}\},
\] a standard prototype on which we do functional analysis, must also have a countable basis. However, if we equip $\ell^\infty$ with a norm $\|(x_n)\|=\sup_{n \ge 1}|x_n|$, then $\ell^\infty$ becomes a Banach space with countable basis.

This is a contradiction to the standard fact in functional analysis that every basis of an infinite dimensional Banach space must be uncountable, so every basis of $\R^\infty$ must be uncountable!