In this paper On prime ideals of Lie algebras, the author proved this theorem:
Let $P$ be an ideal of $L$. Then the following conditions are equivalent:
(i) $P$ is a prime ideal of $L$.
(ii) If $[a,\left\langle b^{L}\right\rangle] \subseteq P$ for $a, b \in L$, then either $a \in P$ or $b \in P$.
(iii) If $[a, H] \subseteq P$ for $a \in L$ and an ideal $H$ of $L$, then either $a \in P$ or $H \subseteq P$.
(i) $\Rightarrow$ (ii). For each $a \in \mathcal{L}$
$$\left\langle a^L \right\rangle= \sum_{i=0}^\infty V_i \qquad \textit{where} \qquad V_0 = (a) \qquad \textit{and} \qquad V_i = \big[\ldots\big[(a),\underbrace{L\big],\ldots,}_i\ L\big].$$
My question is I how can prove that $\sum_{i=0}^\infty V_i$ is an ideal of $L$ and the smallest ideal containing $a$.
i.e
$$\left\langle a^L \right\rangle= \sum_{i=0}^\infty V_i$$
My attempt:
For each $a \in L$, $\langle a^L \rangle$ is defined to be the smallest ideal of $L$ containing $a$. It must contain $a$, and all its scalar multiples, i.e. the $1$-dimensional $F$-vector subspace $(a) := F a \subset \langle a^L \rangle$. But if it contains $(a)$, then (by definition of ideal) it must also contain $[(a),L]$, and hence $[[(a),L],L]$, etc. Because it is a $F$-vector space it must contain their sum.
Therefore, $$\left\langle a^L \right\rangle \supseteq \sum_{i=0}^\infty V_i$$
Conversely, Let $x \in \sum_{i=0}^\infty V_i$ and $y \in L$, then $x \in (a)+[(a),L]+[[(a),L],L]+...$.
Thus $[x,y] \in [(a),L]+[[(a),L],L]+...$.
Therefore, $[x,y] \in \sum_{i=0}^\infty V_i$, hence $\sum_{i=0}^\infty V_i$ is an ideal containing $a$, so it contains $\left\langle a^L \right\rangle$.
Your proof of both inclusions looks good to me. To be very precise, one could make it clear that all actual sums in that infinite sum are actually finite, and mention that $\sum V_i$ is (of course) also a sub-vector space, but I would accept it as written.