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
If $X$ is an infinite-dimensional vector space over some field $F,$ then any basis $B$ must be an infinite set.
It's true that any $v\in X$ can be written as a finite linear combination $p_1 b_1 +\dots + p_n b_n,$ where the $p_k$ are in the underlying field $F$ and the $b_k$ are in the basis $B.$
This doesn't say that $B$ is finite though. Different $v$ in $X$ will require different basis elements to write them in the above format — only finitely many basis elements for any particular $v,$ but (assuming that $X$ is infinite-dimensional so that $B$ is infinite) as you let $v$ vary, you're going to need the infinitely many basis elements to write the various linear combinations (even though each linear combination is itself, individually, a finite sum).
Here's an example:
Let $X$ be the set of all infinite sequences of real numbers that are eventually $0;$ in other words, a member of $X$ is a function $f\colon\mathbb{N}\to\mathbb{R}$ such that for some $n\in\mathbb{N}$, for all $k\gt n,$ $f(k)=0.$ Of course, $X$ is a vector space over $\mathbb{R}$ under pointwise addition and pointwise scalar multiplication.
For each $n\in\mathbb{N},$ let $b_n\in X$ be defined by setting $$b_n(k)=\begin{cases}1,\text{ if }k=n,\\0,\text{ if }k\ne n.\end{cases}$$ Then you can see that $\{b_n\mid n\in\mathbb{N}\}$ is a basis for $X$ over $\mathbb{R}.$ Any member of $X$ can be written as a finite linear combination of the $b_n\text{'s}$ (since each member of $X$ is eventually $0).$ But you need all the $b_n\text{'s}$ (infinitely many) to write all the members of $X$ in that way.
[By the way, the theorem that every vector space has a basis uses the axiom of choice, as does the theorem that any two bases for the same vector space must have the same cardinality. Without the axiom of choice, there can be vector spaces without a basis, and also vector spaces which have a basis but which don't have a well-defined dimension, because different bases can have different cardinalities. I wouldn't worry about any of this when you're just starting to study vector spaces though.]