A coordinate independent proof is possible if you use the definition $L_X = \frac{\mathrm d}{\mathrm d t}\big|_{t=0}\,\Phi_t^*$, where $\Phi$ is the flow of the field $X$, and note that:
- Evaluation $\omega(Y_1,\dots,Y_n)$ of a tensor can be thought of as repeated contraction applied on $\omega\otimes Y_1\otimes Y_2 \otimes\dots\otimes Y_n$.
- If $f : M\to M$ is a diffeomorphism, then $f^*$ commutes with contraction.
- If $f$ is a diffeomorphism, and $A$ and $B$ are arbitrary tensors, then $f^*(A\otimes B) = f^*A\otimes f^*B$.
- $\frac{\mathrm d}{\mathrm d t}\big|_{t=0}(A_t\otimes B_t) = \frac{\mathrm d A_t}{\mathrm d t}\big|_{t=0}\otimes B + A\otimes \frac{\mathrm d B_t}{\mathrm d t}\big|_{t=0}$.
The last part is the only bit where you really have to 'show' something that is related to the Lie derivative and not to general properties of diffeomorphisms, but of course the same trick that works for the Leibniz rule for classical calculus will work here as well.
From 3 and 4 together it follows that
$$ L_X(A\otimes B) = L_X(A)\otimes B + A\otimes L_X(B). $$
So that we have (say for $n=1$):
\begin{aligned} L_X(\omega(Y)) &= L_X(C(\omega\otimes Y)) \\ &= C(L_X(\omega\otimes Y)) \\ &= C(L_X\omega\otimes Y) + C(\omega\otimes L_XY) \\ &= (L_X\omega)(Y)+\omega(L_XY). \end{aligned}
(Where $C$ is a contraction.)
Finally, use the antisymmetrizer and linearity of $L_X$ to obtain the same result for differential forms.
