People keep mentioning the restriction on the size of a Schauder basis, but I think it's more important to emphasize that these bases are bases with respect to different spans.
For an ordinary vector space, only finite linear combinations are defined, and you can't hope for anything more. (Let's call these Hamel combinations.) In this context, you can talk about minimal sets whose Hamel combinations generate a vector space.
When your vector space has a good enough topology, you can define countable linear combinations (which we'll call Schauder combinations) and talk about sets whose Schauder combinations generate the vector space.
If you take a Schauder basis, you can still use it as a Hamel basis and look at its collection of Hamel combinations, and you should see its Schauder-span will normally be strictly larger than its Hamel-span.
This also raises the question of linear independence: when there are two types of span, you now have two types of linear independence conditions. In principle, Schauder-independence is stronger because it implies Hamel-independence of a set of basis elements.
Finally, let me swing back around to the question of the cardinality of the basis.
I don't actually think (/know) that it's absolutely necessary to have infinitely many elements in a Schauder basis. In the case where you allow finite Schauder bases, you don't actually need infinite linear combinations, and the Schauder and Hamel bases coincide. But definitely there is a difference in the infinite dimensional cases. In that sense, using the modifier "Schauder" actually becomes useful, so maybe that is why some people are convinced Schauder bases might be infinite.
And now about the limit on Schauder bases only being countable. Certainly given any space where countable sums converge, you can take a set of whatever cardinality and still consider its Schauder span (just like you could also consider its Hamel span). I know that the case of a separable space is especially useful and popular, and necessitates a countable basis, so that is probably why people tend to think of Schauder bases as countable. But I had thought uncountable Schauder bases were also used for inseparable Hilbert spaces.
Best Answer
The existence of a Hamel basis for $\ell^p$ cannot be proved without some of the axiom of choice, which in modern terms usually means that we cannot write it explicitly.
It is consistent with ZF+DC (a weak form of the axiom of choice which is sufficient to do a lot of the usual mathematics) that all sets of real numbers have Baire property, and in such model we have that every linear function from $\ell^p$ to itself is continuous.
It is also true (in ZF) that $\ell^p$ is separable for $1\leq p<\infty$. It is a known fact that continuous endomorphisms are determined completely by the countable dense set.
If there exists a Hamel basis then its cardinality is at least $\frak c$ (or rather exactly that), and therefore it has $2^\frak c$ many permutations, each extends uniquely to a linear automorphism, which is continuous.
Now, note that $\ell^p$ has size $\leq\frak c$ itself, since it is a separable metric space (and again, this is in fact $\frak c$) and therefore it has only $\frak c$ many continuous endomorphisms.
Cantor's theorem tells us that $2^\frak c\neq c$, and therefore in Shelah's model where every set of real numbers have the Baire property there is no Hamel basis for $\ell^p$.