On the full reducibility of representations of reductive Lie algebras I would like to find a reference for the following fact:
every finite dimensional complex representation of a reductive Lie algebra is semisimple.
 A: You can find the proof in Serre's "Lie Algebras and Lie Groups", in chapter "Semisimple Lie Algebras", section "Complete Reducibility"
A: To complement Jim's answer, there is a thorough discussion of complete reducibility for reductive Lie algebras (with proofs, but only in char=0) in Sections 1.6 and 1.7 of Dixmier's "Enveloping algebras", which I found much less intimidating then reading Bourbaki.
A: In many applications, a (real) reductive Lie algebra arises as the Lie algebra of a compact Lie group.  In this case, and if the representation integrates to one of the group, then it is fully reducible by a version of Weyl's unitary trick.  Basically every finite-dimensional module is unitarisable and every submodule has a complementary submodule: namely, its perpendicular complement.
A: The statement is false.   The standard definition of "reductive" for a finite dimensional Lie algebra $\mathfrak{g}$ over an arbitrary field of characteristic 0 is given in a number of equivalent ways by Bourbaki in Chapter 1 (1960) of their treatise on Lie groups and Lie algebras: section 6, no. 4-5.   By definition, $\mathfrak{g}$ is reductive provided its adjoint representation is semisimple (= completely reducible). Typical equivalent conditions: the derived algebra is semisimple; or $\mathfrak{g}$  is the direct sum of a semisimple and an abelian Lie algebra; or the solvable radical equals the center.
As a consequence, a finite dimensional representation of a reductive Lie algebra is semisimple iff the center acts by semisimple endomorphisms.   (An abelian Lie algebra need not be represented in that way.)
Some of this is set up as an exercise at the end of Section 6 in my Springer graduate text (1972); see also Proposition 19.1.
