Let $G$ be a graph with disjoint copies of $K_{1,3}$. Prove that if there are uncountably many copies of $K_{1,3}$ in $G$, then $G$ is not planar.

I have a proof of this statement by contradiction i.e. assuming it is planar with uncountably many copies of $K_{1,3}$ in $G$, but I am not satisfied. I am wondering if anyone has a proof that proves the contrapositive i.e. if $G$ is planar, then there is only countably many copies of $K_{1,3}$ in $G$.

My idea is to find an embedding of $G$ onto $\mathbb{R}^2$ (or $\mathbb{C}$), and use the fact that $K_{1,3}$ has $3$ edges, and hence if we find disjoint open sets which contain each $K_{1,3}$, they will have positive measure.

Thanks a lot!