Is the cardinality of the basis of a vector space $V$ having infinite dimension necessarily countable infinite? 
Suppose a vector space $V$ has infinite dimension, $\dim V = \infty$. Is the cardinality of the basis of $V$ neccesarily countable infinite?

If a vector space has infinite dimension does $V$ then have a basis ? If one can specify a set of infinitely many linearly independent vectors is this set then regarded as a basis for $V$ ?
Are there results that can be proven for vector spaces with finite dimension, but not for vector spaces with infinite dimension ?
Why does most books on linear algebra only mention finite dimension vector spaces ?
 A: *

*Not all infinite dimensional vector spaces have countable basis. In fact, most spaces you come across in "real life" don't. The simplest example would be a sequence space like
$$
\ell^2 = \{(x_n) \subset \mathbb{C} : \sum_{n=1}^{\infty} |x_n|^2 < \infty\}
$$
This is an example of a Banach space - a vector space that comes with a nice notion of distance (a norm) and one that is complete with respect to that norm.
It is a general fact that an infinite dimensional Banach space cannot have a countably infinite basis (due to the Baire Category theorem).

*However, the space
$$
c_{00} := \{(x_n) \in \mathbb{C} : x_n \neq 0 \text{ for only finitely many } n\}
$$
is an example of an infinite dimensional vector space whose basis is countable (the "standard" basis). 

*Most books on Linear Algebra mention only finite dimensional vector spaces because they are easy to visualize (just extend your notion of a vector in $\mathbb{R}^2$), but they are also deep enough to prove some rather interesting results (for instance the spectral theorem). Furthermore, infinite dimensional vector spaces are best analysed with some topology in mind - this is what functional analysis studies.
A: Let $X$ be any set with cardinality $\alpha$, then consider the set of formal linear combinations $$\{a_1x_1+\cdots+a_nx_n: a_i\in K, x_i \in X, n\in\mathbb{N}\}$$
When equipped with usual addition and multiplication, this set forms a vector space with basis $X$ and dimension $\alpha$. So we have vector spaces of arbitrary dimensions.
A: If you accept the axiom of choice, then every vector space has a basis. This  result can be further improved to $R$-modules over division rings I guess.
Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. What's the cardinality of a basis for $\mathbb{R}$ over $\mathbb{Q}$? Think about it.
Think of a basis as a maximal linearly independent set. Can you somehow use Zorn's lemma to prove that every vector space has a basis now?
Why does most books on linear algebra talk about only the finite dimensional case? Because the infinite dimensional case requires some knowledge of Analaysis I guess. And most tools that we employ in linear algebra work because we have matrices. We don't have such tools in the infinite-dimensional case, so we need to extend our tools and theory.
Addendum: Read this page. It should answer a lot of your questions.
