Given a subspace $V$ of a metric Lie algebra, does $[[V, V], V] \subset V$ imply $[V, V] \subset V$?

Let $\mathfrak{g}$ be a (finite-dimensional, real) Lie algebra equipped with a nondegenerate symmetric bilinear form $\langle \cdot{,}\cdot \rangle$ that is ad-invariant, i.e., such that \begin{equation*} \langle [x,y],z\rangle = \langle x,[y,z]\rangle \quad \text{for all $x,y,z \in \mathfrak{g}$}; \end{equation*} let $V \subset \mathfrak{g}$ be a nondegenerate subspace.

Question. Suppose that $[[V, V],V] \subset V$, meaning that $[[u, v], w] \in V$ for all $u, v, w \in V$. Does it then follow that $V$ is a Lie subalgebra, i.e., that $[V, V] \subset V$?

Some thoughts: By the ad-invariance of the metric, \begin{equation*} \langle [u, v], [w, \eta] \rangle = \langle [[u, v], w], \eta \rangle = 0 \quad \text{for all $u,v,w \in V$ and $\eta \in V^{\perp}$}, \end{equation*} implying $[V, V^{\perp}] \subset [V, V]^{\perp}$.

Take $\mathfrak{g} = \mathbb{R}^3$ with $[x,y] = x \times y$ the cross product and as symmetric bilinear form $\langle \cdot, \cdot \rangle$ the usual scalar product. Then the space $V = \langle e_1, e_2 \rangle$ is not closed under the Lie bracket since $[e_1,e_2] = e_1 \times e_2 = e_3$. However, since we have $[V,V] = \langle e_3 \rangle$, we find that $[[V,V],V] \subseteq V$ even though $V$ is not a Lie subalgebra.

My answer to another of your questions also applies here. Take a Cartan decomposition of a semisimple Lie algebra or more generally a symmetric decomposition.

A symmetric decomposition is defined as $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}$ where $[\mathfrak{h}, \mathfrak{h}] \leq \mathfrak{h}$, $[\mathfrak{h}, \mathfrak{m}] \leq \mathfrak{m}$ and $[\mathfrak{m}, \mathfrak{m}] \leq \mathfrak{h}$. It is a quick check to see that a Cartan decomposition is symmetric. More generally these correspond to symmetric homogeneous space so you can find a whole load of examples of these.

Immediately you get $[\mathfrak{m}, [\mathfrak{m}, \mathfrak{m}]] \leq \mathfrak{m}$ and this breaks your rule.