Show that if a Banach space is generated by a countable set with finite dimension then it has finite dimension I don't know what to do with the following question:

Show that if a Banach space is generated by a countable set with finite dimension then it has finite dimension

Supposedly it's related to the Baire theorem, but that didn't help me one bit. In fact it looks totally unrelated:

Theorem (Baire) If $(G_n)_{n \geq 1}$ is a sequence of open, dense subsets of the complete metric space $(X,d)$, then the intersection $\bigcap_{n \geq 1} G_n$ is dense in $X$.

So how would one go solving this?
 A: Edit At first I hadn't read the title of your question so I thought that you wanted to prove Baire's theorem and I added an answer based on that. My bad.
However, I do know how to show what you have on your title using Baire's theorem so here's a sketch:
I'll leave to you to check all the details.
The following statement is a direct consequence of Baire's theorem

Theorem. If $(X,d)$ is a complete metric space and $X=\bigcup_{n \geq 1} X_n$, then $\mathrm{int}(\overline{X_n}) \neq \varnothing$ for some $n$.

Now I will use the theorem to show that the algebraic dimension of a Banach space is either finite or uncountable:
Assume that $X$ is a Banach space generated by $\{ \xi_n : n \in \mathbb{Z}_{>0}\}$. For each $n \in \mathbb{Z}_{>0}$ let $X_n$ be the space generated by $\{\xi_1, \ldots, \xi_n\}$. Now it's standard to show that

*

*$X_n$ is closed

*$\mathrm{int}(X_n)= \varnothing$
Since $X$ is generated by these $X_n$'s, we have $X=\bigcup_{n \geq 1} X_n$. Thus, by the theorem above $\mathrm{int}(X_n) \neq \varnothing$ for some $n$, a contradiction.
