About finite dimensional vector spaces Would you tell me why the statement below holds? 

A vector space $V$ has a basis if and only if $0 < \dim V < \infty.$

 A: The claim is false, whether or not one accepts the Axiom of Choice.
Consider the vector space $V$ (over the reals) of all polynomials $P(x)$ with real coefficients. Addition of polynomials, and multiplication by a scalar, are defined in the usual way.  
The space $V$ is infinite-dimensional, but has basis $\{1,x,x^2,x^3,x^4,\dots, x^n, \dots\}$.
Remark: It is not difficult to show, without using the Axiom of Choice, that any finite dimensional space does have a basis, so the implication in one direction is true. 
A: Your claim is not true. There certainly exists infinite dimensional vector spaces. 
However, the statement that every vector space has a basis is provable depending on your axioms. 
For example,  $ZF$ without foundation and without the power set axiom but with the well-ordering principle can prove that every vector space has a basis. 
Moreover, ZF with the fact that every vector space has a basis can prove the axiom of choice (and hence the well-ordering principle). See Existence of Bases Implies the Axiom of Choice by Blass. This show that over $ZF$, the axiom of choice is equivalent to the fact that every vector space has a basis. 
Also the axiom of choice is independent of $ZF$, so using just $ZF$ alone you cannot prove that every vector space has a basis. 
