= Solution
A <Cartan subalgebra> is a nilpotent <Lie subalgebra> equal to its <normalizer of a Lie subalgebra>. Write $H=L_{0,x}$ for a choice $x\in A$ attaining the minimum dimension, and let $h=\dim H$. For $y\in A$, the subalgebra $H$ is invariant under $\operatorname{ad}(x+ty)$ because $A\subseteq H$. On $L/H$, the operator induced by $\operatorname{ad}x$ is invertible: $H$ is its full generalized zero eigenspace, and the <finite-dimensional Fitting decomposition> separates that from the invertible part.
The determinant of the operator induced by $\operatorname{ad}(x+ty)$ on $L/H$ is a <polynomial> in $t$, nonzero at zero. Thus it is nonzero for all but finitely many $t\in K$. For these values, the generalized zero eigenspace $L_{0,x+ty}$ lies in $H$. Minimality of $\dim H$ among elements from $A$ forces equality. Hence $\operatorname{ad}(x+ty)|_H$ is nilpotent for all but finitely many $t$.
Each nonleading coefficient of its <characteristic polynomial> is a <polynomial> in $t$. Since $K$ is infinite, a <polynomial> vanishing at all but finitely many elements is identically zero. The coefficient of degree $i$ in the <matrix> entries is homogeneous of degree $i$; its highest possible $t^i$ coefficient is the corresponding characteristic coefficient of $\operatorname{ad}y|_H$. All these vanish, so that operator has <characteristic polynomial> $T^h$ and is nilpotent by the <Cayley-Hamilton theorem>. Thus
$$
\boxed{H\subseteq L_{0,y}\qquad(y\in A).}
$$
For existence, choose an $L_{0,x}=H$ of smallest dimension among all $x\in L$. Since $x\in H$, it also minimizes the same set with $x$ restricted to $A=H$. Applying the proved statement makes every $\operatorname{ad}y|_H$ nilpotent. <Engel's theorem> makes $H$ nilpotent, and the allowed self-idealizer property makes it self-normalizing. Therefore it is a <Cartan subalgebra>, illustrating <Cartan subalgebra existence from a minimal Engel subalgebra>.
For the three-dimensional example, direct use of the brackets gives $C_L(z)=Kz$ for every nonzero $z$: the equations $[u,z]=0$ are the vanishing of all two-by-two minors of the two coordinate <vectors>, so they are proportional. Every nonzero nilpotent <Lie algebra> has a nonzero center, as seen from its last nonzero lower-central term. A nilpotent subalgebra $H$ with a central nonzero element $z$ must therefore satisfy $H\subseteq C_L(z)=Kz$, and has dimension one. The zero subalgebra is not self-normalizing. Thus \b[all <Cartan subalgebras> have dimension one]. For example $Ka$ is nilpotent and its normalizer is $Ka$, so such subalgebras do occur.
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