Let $\mathbb{X}$, $\mathbb{Y}$ be vector fields on $U \subset \mathbb{R}^n$. Prove that
$$L_{\mathbb{X}}i_\mathbb{Y}=i_{[\mathbb{X},\mathbb{Y}]}+i_\mathbb{Y}L_{\mathbb{X}} $$
using induction. Assume the following identity $L_{\mathbb{X}}L_{\mathbb{Y}}-L_{\mathbb{Y}}L_{\mathbb{X}}=L_{[\mathbb{X},\mathbb{Y}]}$.
This is clearly true for $0$-forms. If we take $f \in \Lambda^{0}(\mathbb{R}^n)$. Then we get $0=0+0$.
Assume true for $(k-1)$-forms. Then for $k$-forms as $i_{\mathbb{X}}$ and $L_{\mathbb{X}}$ are linear we may just consider $\omega \in \Lambda^{k}(\mathbb{R}^n)$ such that $\omega = f \ d\alpha$.
I rewrote $L_{\mathbb{X}}i_\mathbb{Y}-i_\mathbb{Y}L_{\mathbb{X}}=i_{[\mathbb{X},\mathbb{Y}]}$. As I believe this may be easier to work with.
I cannot see how to incorporate $L_{\mathbb{X}}L_{\mathbb{Y}}-L_{\mathbb{Y}}L_{\mathbb{X}}=L_{[\mathbb{X},\mathbb{Y}]}$, not sure if it would be best to break it down with Cartan's formula. Also I cannot see how to use the inductive hypothesis.
Inductive proofs only please. Note that $[\cdot,\cdot]$ is the Jacobi bracket and no use of the Poisson bracket.
As you wrote in your question, you may consider a $k$-form $\omega$ which is of the form $\omega=fd\alpha$ for a $(k-1)$-form $\alpha$.
Step 1: Clearly, $i_{[ \mathbb{X}, \mathbb{Y}]} fd\alpha= f i_{[\mathbb{X},\mathbb{Y}]} d\alpha$. On the other hand, $$ L_{\mathbb{X}} i_\mathbb{Y} fd\alpha = L_{\mathbb{X}} f i_\mathbb{Y} d\alpha = (\mathbb{X}\cdot f) i_\mathbb{Y} d\alpha+fL_{\mathbb{X}} i_\mathbb{Y} d\alpha, $$ and $$ i_\mathbb{Y} L_{\mathbb{X}}fd\alpha = i_\mathbb{Y}(\mathbb{X}\cdot f)d\alpha +i_\mathbb{Y} f L_{\mathbb{X}}d\alpha = (\mathbb{X}\cdot f) i_\mathbb{Y} d\alpha + fi_\mathbb{Y} L_{\mathbb{X}}d\alpha. $$ Taking the difference of these two equations, we see that $$ (L_{\mathbb{X}} i_\mathbb{Y}-i_\mathbb{Y} L_{\mathbb{X}})fd\alpha=f(L_{\mathbb{X}} i_\mathbb{Y}-i_\mathbb{Y} L_{\mathbb{X}})d\alpha. $$ Hence the claim for $\omega=fd\alpha$ will follow if we verify equality for $d\alpha$.
Step 2: To verify the claim fo $d\alpha$, we observe that using Cartan's formula we get $L_{\mathbb{X}} i_\mathbb{Y} d\alpha=L_{\mathbb{X}} L_\mathbb{Y}\alpha-L_{\mathbb{X}} d i_\mathbb{Y} \alpha$, and the last term can be written as $-d L_{\mathbb{X}} i_\mathbb{Y} \alpha$. Similarly, we get $ i_\mathbb{Y}L_{\mathbb{X}} d\alpha=i_\mathbb{Y} d L_{\mathbb{X}} \alpha= L_\mathbb{Y} L_{\mathbb{X}} \alpha - d i_\mathbb{Y} L_{\mathbb{X}} \alpha$. Subtracting these two terms from each other, we obtain $$ (L_{\mathbb{X}} i_\mathbb{Y} - i_\mathbb{Y}L_{\mathbb{X}}) d\alpha=(L_{\mathbb{X}} L_\mathbb{Y}-L_{\mathbb{Y}} L_\mathbb{X})\alpha - d(L_{\mathbb{X}} i_\mathbb{Y}- i_\mathbb{Y} L_{\mathbb{X}})\alpha. $$ By the formula for Lie derivatives you mention in your question, the first term in the right hand side equals $ L_{[\mathbb{X},\mathbb{Y}]}\alpha$. By induction hypothesis, the second term in the right hand side equals $di_{[\mathbb{X},\mathbb{Y}]}\alpha$, so overall we get $$ (L_{\mathbb{X}} i_\mathbb{Y} - i_\mathbb{Y}L_{\mathbb{X}}) d\alpha= L_{[\mathbb{X},\mathbb{Y}]}\alpha - di_{[\mathbb{X},\mathbb{Y}]}\alpha,$$ which by Cartan's formula equals $i_{[\mathbb{X},\mathbb{Y}]}d\alpha$. This completes the proof.