Let $U\in U(H)$ be unitary operator. Is it possible for it to has an empty point spectrum?
I am aware that every bounded operator acting on complex Hilbert space has non-empty spectrum.
Since $\forall \lVert v \rVert = 1$ we have
$$\lVert Uv \rVert ^2 = \langle Uv, Uv\rangle = \langle U^*Uv, v \rangle = \lVert v \rVert ^2 = 1$$
It means that indeed $\sigma{(U)} \neq \emptyset$.
The question is how can I prove such thing without definition of resolvent and Liouville's theorem?
Yes, there are many well-known unitaries with empty point spectrum. For instance consider $H=L^2(\mathbb T)$, with Lebesgue measure, and $U$ the operator given by $$ (Uf)(z)=zf(z). $$ As $|z|=1$ for all $z$, it is easy to see that $U$ is isometric and that $U^*U=UU^*=I$. It is easy to check that $\sigma(U)=\mathbb T$, and that it has no eigenvalues.
The argument you give to claim that any unitary has non-empty spectrum does not prove anything. You could have used it to prove the same over a real Hilbert space, where it is not true: the unitary $$ U=\begin{bmatrix} 0&1\\-1&0\end{bmatrix} \in B(\mathbb R^2) $$ has empty spectrum. So to prove that the spectrum is nonempty, you have to use that your field is $\mathbb C$ in an essential way. You will not be able to avoid either the Fundamental Theorem of Algebra, Liouville, or some other similar theorem.