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.
Your definitions of "linearly independent", "basis", and "orthonormal basis" are all correct. In particular, an orthonormal basis for an infinite-dimensional Hilbert space is not actually a basis (since you will need to use infinite linear combinations).
"Hamel basis" means exactly the same thing as "basis". The reason that it is given a different name is to emphasize that you are talking about a basis with respect to finite linear combinations, as opposed to some other kind of object that might be referred to using the word "basis" but which is not actually a basis (such as an orthonormal basis or a Schauder basis). Indeed, when talking about infinite-dimensional topological vector spaces, it is rare that you actually care about a basis as opposed to some related notion that allows for infinite linear combinations. So in most contexts if someone refers to a "basis" of such a space, it is actually more likely than not that they are abusing terminology and are using "basis" as an abbreviation for "orthonormal basis" or something similar. To make it clear that you literally mean just a basis, it is common to say "Hamel basis".
As for your book's definition, note that "maximal" means "cannot be enlarged to a superset", not "cannot be enlarged in cardinality". So a "maximal linearly independent set" is a linearly independent set $S$ such that there is no proper superset $T$ of $S$ which is linearly independent. This is equivalent to saying $S$ spans the whole space (using finite linear combinations). Indeed, if $S$ does not span the whole space, you can take any vector not in its span and add it to $S$ to get a larger linearly independent set. Conversely, if $S$ does span the whole space, any vector not in $S$ is a linear combination of elements of $S$ and thus would give a linearly dependent set if you added it to $S$.
Best Answer
If $n\in\Bbb N$,$$x^n=\bigl((x-1)+1\bigr)^n=\sum_{k=0}^n\binom nk(x-1)^k,$$and therefore $x^n$ is a linear combination of polynomials of the form $(x-1)^k$. So, since the $x^n$'s span $\Bbb Q[x]$, then $(x-1)^n$'s also span it.
Now, let us see that $\{1,x-1,(x-1)^2,\ldots,(x-1)^n\}$ is linearly dependent. Let $a_0,a_1,\ldots,a_n$ be scalars such that$$a_0+a_1(x-1)+a_2(x-1)^2+\cdots+a_n(x-1)^n=0.$$If you expand the LHS of the previous equality, you get a polynomial of degree $n$ such that the coefficient of $x^n$ is $a_n$. But the $x^n$'s are linearly independent, and therefore $a_n=0$. So, you know that$$a_0+a_1(x-1)+a_2(x-1)^2+\cdots+a_{n-1}(x-1)^{n-1}=0,$$and you can start all over again, showing that $a_{n-1}=a_{n-1}=\cdots=a_1=a_0=0$.