Linear space with (Hamel) basis and the axiom of choice It is true that the axiom of choice is equivalent to the statement that every linear space has a Hamel basis. There are some linear spaces which definitely don't need axiom of choice to possess (rather canonical) basis: for example $c_{00}$, the space of all sequences with compact supports. Is it possible to give an example of one linear space $V$ with the property: the existence of basis in $V$ implies the axiom of choice?
 A: No. It is not possible.
Suppose that $V$ is a specified vector space, then it is consistent that the axiom of choice fails very far above $V$ in the hierarchy of sets (the von Neumann hierarchy). In particular it would mean that $V$ has a basis, but still the axiom of choice fails, as it fails very far above $V$.
Generally speaking, no particular set can witness the axiom of choice. It is possible, that one set can decide the axiom of choice (so it is possible that one vector space's basis decides the axiom of choice), but this requires additional assumptions. More specifically, this assumption was formulated by Andreas Blass as "Small Violations of Choice" (SVC). It means that there is a set which "more or less" decides a lot of the information regarding the axiom of choice in the universe. From this set we can create a vector space, that has a basis if and only if the original set can be well-ordered and that would imply the axiom of choice in full.
A: Yes, it is true that AC is equivalent to the assertion that every vector space has a basis, and this is discussed in all the usual treatments of equivalents to the axiom of choice. For example, the reference is given on the wikipedia entry for the axiom of choice. The result is due to Andreas Blass, who is active here on MathOverflow. 
Regarding your request for a single space witnessing AC, there is a sense in which this is trivially true. If AC fails, then let $V$ be any particular space without a basis. Then, for this particular space, it is true to say:


*

*If $V$ has a basis, then AC holds.


And furthermore, if AC holds, then any space $V$ makes that statement true. But I realize that is not what you meant. 
Meanwhile, as Asaf points out, there is a general sense in which no particular well-order is sufficient to ensure AC, because there could always be higher violations of AC at higher cardinals. The symmetric model methods allow you to build models of $\neg\text{AC}$ while preserving AC in any desired rank-initial segment of the universe. So that is one way of answering your final request negatively.
