The set of real sequences has no countable spanning set I'm working through some old Harvard Math 55 problem sets, and one problem in particular has got me stumped:

If $F$ is field, then prove that the vector space $F^\infty$ over $F$ (the space of infinite tuples or sequences) has no countable spanning set. 

In the case that $F$ is countable or finite, we can show that the space spanned by a countable set is necessarily itself countable and therefore cannot span $F^\infty$, which is uncountable. However, that argument fails when the field is uncountable, since the span of even a single vector is an uncountable set. Any hints on how to proceed?
 A: Here is one idea/hint for a proof by diagonalization. Given a countable set of sequences $C$, we want to make a sequence $s$ that is not in the span of $C$, which is to say $s$ is not a linear combination of any finite subset of $C$.  
To do this, it is sufficient to make an $s$ such that, for each finite subset $T$ of $C$, say of size $k_T = k$, there are some $k+1$ coordinates of $s$ such that the vectors we get in $\mathbb{R}^{k+1}$ by restricting $T$ to those coordinates do not span the vector in $\mathbb{R}^{k+1}$ obtained by restricting $s$ to those coordinates. This guarantees that $s$ is not in the span of $T$.
No set of $k$ vectors in $\mathbb{R}^{k+1}$ can span $\mathbb{R}^{k+1}$, so if we choose a set of new coordinates for each finite subset $T$ of $C$, we can choose values of $s$ on those coordinates appropriately to ensure the previous condition holds. This means we just need to set up the appropriate construction to make $s$. 
Separately, there is a more well known argument that shows that no infinite dimensional Banach space can have a countable basis; see the question Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable. In this case, we only have a vector space, so the techniques used there don't seem to apply. 
