One of the definitions of Cartan subalgebra $\mathfrak{h}$ of a semisimple Lie algebra $\mathfrak{g}$ one can find in the literature is that it is a
maximal abelian subalgebra
has the property that $ad_h$ is diagonalizable for all $h\in\mathfrak{h}$
I wonder whether, at least over an algebraically closed field of characteristic zero, the condition 2 is vacuous?
It is not vacuous. Consider $\mathfrak{g} = \mathfrak{sl}_2(\Bbb C)$, $\mathfrak{h} = \Bbb C x$ for any non-semisimple element $x\in \mathfrak{g}$ (for example, $\pmatrix{0&1\\0&0}$): It satisfies 1, but not 2, and is not a Cartan subalgebra.
Maybe it is noteworthy that a subalgebra which consists of $ad$-diagonalisable elements is automatically abelian (cf. Humphreys' Introduction to Lie Algebras and Representation Theory, section 8.1; see also here); so as long as you are over an algebraically closed field (of char. $0$), you can instead reduce the definition to "maximal with the property that all its elements are $ad$-diagonalisable". But as soon as the ground field is not algebraically closed, so that the notions of "semisimple element" and "$ad$-diagonalisable element" become distinct, this can fail badly. It would still characterise maximal split toral subalgebras, but not maximal toral algebras, which are the Cartan subalgebras. For that distinction and examples, cf. my blunder in Characterisation of Cartan subalgebras as maximal toral; redundant "abelian" in the definition of "toral", and the correct approach in https://math.stackexchange.com/a/2497093/96384.
