The mapping $(A,B)\mapsto A\otimes B$ is a bi-linear mapping from $\Hom(E)\times\Hom(F)$ into $\Hom(E\otimes F)$ and thus, by the universal property there is a linear mapping $\Phi:\Hom(E)\otimes\Hom(F)\rar\Hom(E\otimes F)$ such that $\Phi(A\otimes B)=A\otimes B$ - here the tensor product on the left hand side is our 'old' definition of the tensor product of two spaces and on the right hand side it's the tensor product of operators. Since $E,F$ are finite dimensional and $\Phi$ maps a basis onto a basis, $\Phi$ must be an isomorphism.