In Milne's notes on Lie algebras, part (a) of theorem 5.20 (p52) states:
Let $\mathfrak{g}$ be a Lie algebra over $k$. Suppose the adjoint representation $\mathfrak{g}\rightarrow \mathfrak{gl}_{\mathfrak{g}}$ is semisimple, then $\mathfrak{g}$ is semisimple.
His proof goes:
If the adjoint representation is semisimple, then $\mathfrak{g}$ is a direct sum of ideals, and every ideal admits a complementary ideal, so is a quotient of $\mathfrak{g}$. Thus, if $\mathfrak{g}$ is not semisimple, then it has a commutative ideal, and hence a commutative quotient, and hence a 1-dimensional quotient. However, 1-dimensional Lie algebras have non-semisimple representations.
I agree with everything he says, but I don't follow his logic. Why does having a 1-dimensional quotient prevent the adjoint representation from being semisimple?
Yes, you are right, this statement is not true. I think he has a copy-paste error here. In fact, we should read Proposition $6.2$ of Milne's lecture notes, on page $57$. There he shows that $\mathfrak{g}$ is reductive if and only if the adjoint representation of $\mathfrak{g}$ is semisimple, if and only if $\mathfrak{g}$ is a product (direct sum) of a semisimple Lie algebra and a commutative one.
Note: Milne himself has acknowledged the error on his web page here: From David Calderbank Page 52, Weyl's Theorem 5.20. Part (a) states "If ad is semisimple, then g is semisimple". However, semisimplicity of the adjoint representation is characteristic of reductive Lie algebras (Proposition 6.2), not just semisimple ones. The proof uses more than semisimplicity of the adjoint representation.