Define $V$ as the subset of $\mathbb{R}[[x]]$, the $\mathbb{R}$-vector space of formal power series with real coefficients, such that for any $f\in V$, for any $r\in\mathbb{R}$, the series $f$ always converges when evaluated at $r$. It can be shown that $V$ is a subspace of $\mathbb{R}[[x]]$.
I am trying to show that $V$ does not have a countable basis. Suppose for contradiction that $S$ is a countable basis of $V$. Write the $i$th element in $S$ as $f_i:=\sum_j a_{ij}x^j$. One attempt to construct an element $g$ in $V$ such that $g$ is not spanned by $S$ would be some kind of diagonal argument, i.e. $g:=\sum_j h(a_{jj})x^j$ for some suitable function $h$. However, I struggle to find an $h$ such that $g$ is convergent because the $a_{jj}$ can be quite arbitrary. Is there any other promising way I can do this?
Consider the subspace $U\subset V$ consisting of all functions of the form $$f(x)=\sum_{n=0}^\infty {c_n\over n!}x^n$$ where $c=\{c_n\}\in\ell^\infty.$ The space $U$ is isomorphic to $\ell^\infty,$ hence it does not admit a countable basis, as an infinite dimensional Banach space. Therefore $V$ does not admit such a basis.