A Cartan subalgebra of an arbitrary Lie algebra means a maximal abelian subalgebra

The subalgebra $\mathbb{C}Y$ is maximal abelian: it is one-dimensional, hence abelian, and $\mathfrak{g}$ itself is not abelian, so no abelian subalgebra strictly contains it.

But $\mathbb{C}Y$ is not a Cartan subalgebra: $N_{\mathfrak{g}}(\mathbb{C}Y)=\{aX+bY:[aX+bY,Y]\in \mathbb{C}Y\}$ equals $\mathfrak{g}$, because $[X,Y]=Y\in \mathbb{C}Y$ and $[Y,Y]=0$; a Cartan subalgebra would have to equal its normalizer, and $\mathbb{C}Y\mathrm{\ne }\mathfrak{g}$.

The definition is nevertheless not vacuous: $\mathbb{C}X$ is a Cartan subalgebra of $\mathfrak{g}$, since $[aX+bY,X]=-bY\in \mathbb{C}X$ forces $b=0$, so $N_{\mathfrak{g}}(\mathbb{C}X)=\mathbb{C}X$, and $\mathbb{C}X$ is abelian and therefore nilpotent. Thus a maximal abelian subalgebra of an arbitrary Lie algebra need not be a Cartan subalgebra, and the proposed identification fails.