I read the fact that if $\mathfrak{h}$ is an abelian Lie algebra, and $V=\bigoplus_{\lambda\in\mathfrak{h}^\ast} V_\lambda$ is an $\mathfrak{h}$-module, then for any $\mathfrak{h}$-submodule $U$ of $V$, $U$ decomposes as $U=\bigoplus_{\lambda\in\mathfrak{h}^\ast}(U\cap V_\lambda)$. Here $V_\lambda=\{v\in V: h\cdot v=\langle h,\lambda\rangle v,\forall h\in\mathfrak{h}\}$.
Proof: For $u\in U$, write $u=v_1+\cdots+v_m$ for $v_i\in V_{\lambda_i}$, and $\lambda_1,\dots,\lambda_m$ distinct. There is $h\in\mathfrak{h}$ such that $\langle h,\lambda_1\rangle,\dots,\langle h,\lambda_m\rangle$ are all distinct.
We have a system of equations $$ h^i\cdot u=\langle h,\lambda_1\rangle^iv_1+\cdots\langle h,\lambda_m\rangle^iv_m $$ for $i=0,\dots,m-1$. This is a system in $v_1,\dots,v_m$ whose coefficient matrix is a Vandermonde matrix, hence is invertible since the $\langle h,\lambda_i\rangle$ are all distinct. So $v_i$ can be solved in terms of the $h^i\cdot u$, hence is in $U$, giving the decomposition.
Where is the hypothesis that $\mathfrak{h}$ be abelian used?
Let $V$ be a $\mathfrak h$-module, where $\mathfrak h$ might not be abelian. That is, an Lie algebra homomorphism $\mathfrak h \to \mathfrak{gl}(V)$. Then we have for all $v\in V$, $h_1, h_2 \in \mathfrak h$,
$$[h_1, h_2]v = h_1 (h_2 v) - h_2 (h_1 v)$$
Now if we use the definition of $V_\lambda$, this means
$$[h_1, h_2] v = 0\ \ \ \ \forall h_1, h_2 \in \mathfrak h, v\in V_\lambda.$$
Thus the Lie algebra homomorphism $\mathfrak h \to \mathfrak{gl}(V)$ descends to $\mathfrak h/I \to \mathfrak{gl}(V)$, where $I$ is the commumtator ideal (the ideal generated by $[h_1, h_2]$). Note that then $\mathfrak h/I$ is abelian.
Thus, that $\mathfrak h$ is abelian is almost used when one defines $V_\lambda$.