linear algebra - Exact short sequences of vector spaces

Without using any basis of A, B or C specifically, but rather using the existence of bases for all vector spaces, we may observe that all vector spaces are free as modules; and all free modules are projective.

Finally, the projectivity property gives us a splitting by lifting the identity map on C along the surjection from B by projectivity of C. This gives us a splitting, thus splitting the entire sequence.

ETA With the arguments in the comments to Ben Websters answer, it is pointed out that without AC, and without at least the existence of bases, things can fail badly. Obviously, if we disallow AC, this answer fails as badly.