Let $V$ be a vector space, $S\subseteq V$ a linearly independent subset and $\mathcal{A}=\{T\subseteq V: S\subseteq T \text{and $T$ is linearly independent}\}$. It is easy to see that any chain on $\mathcal{A}$ has an upper bound on $\mathcal{A}$ (we can take the union). Then, it follows from Zorn's lemma that $\mathcal{A}$ has a maximal element $R$. If $\langle R\rangle\neq V$ then we can consider $R\cup\{v\}$ for some $v\notin \langle R\rangle$ and we obtain an element of $\mathcal{A}$ which is greater than a maximal element. The contradiction comes from our assumption that $\langle R\rangle\neq V$. So, we must have $\langle R\rangle = V$.
Let me answer your last two questions first:
In the example of $Q[t]$, what I would call the "standard basis" is $\{1, t, t^2, t^3, \dots \}$. This is a basis because every polynomial in $Q[t]$ can be uniquely expressed as a linear combination of those elements. But there are many other (in fact, infinitely many) bases for $Q[t]$; for example, $\{1, 1+ t, 1 + t +t^2, 1 + t + t^2 + t^3, \dots \}$ is also a basis because, again, every polynomial can be uniquely expressed as a linear combination of those elements. (As an exercise, can you express, say, $t^2 +3t -1$ as a linear combination of the elements in that basis?) Remember, vector spaces generally have many different bases.
Now on to your first question. As other answers have said, every vector space $V$ has a basis, which is a set of vectors $B$ such that every vector in $V$ can be uniquely written as a linear combination of vectors in $B$. (Note: by definition, linear combination means a finite sum.)
Unfortunately, for a lot of commonly occurring infinite dimensional spaces (e.g., function spaces such as the one you mentioned), there's no good way of writing down an explicit basis.
One last note: when one studies Hilbert spaces (which are "complete inner product spaces"), one uses the term "basis" (or "orthonormal basis") to mean something subtly different from our use of "basis" above. In that context, you are allowed to take infinite sums of "basis" elements. This confused me when I first studied Hilbert spaces.
Best Answer
It's known that the statement that every vector space has a basis is equivalent to the axiom of choice, which is independent of the other axioms of set theory. This is generally taken to mean that it is in some sense impossible to write down an "explicit" basis of an arbitrary infinite-dimensional vector space. On the other hand,