Complexity Classes UP and US
Let me start with your third question. It seems that you don't quite understand the definitions of $\mathsf{UP}$ and $\mathsf{US}$, so let me repeat them here.

Definition of $\mathsf{UP}$. Property U. A non-deterministic Turing machine has property U if every input has at most one accepting path. The language accepted by such a machine consists of all inputs which do have an accepting path.
The class $\mathsf{UP}$ consists of all languages accepted by non-deterministic Turing machines having property U and running in polynomial time.
Definition of $\mathsf{US}$. U-acceptance. The language accepted by a non-deterministic Turing machine consists of all inputs which have an accepting path. In contrast, the language U-accepted by a non-deterministic Turing machine consists of all inputs having exactly one accepting path. Any input having either no accepting path or at least two accepting path does not belong to the language U-accepted by the machine.
The class $\mathsf{US}$ consists of all languages U-accepted by non-deterministic Turing machines running in polynomial time.

As for your second question, to show that $\mathsf{coNP} \subseteq \mathsf{US}$ it suffices to reduces $\mathsf{coSAT}$ to $\mathsf{US}$. Given a SAT instance $\phi = C_1 \land \cdots \land C_m$, let
$$\psi = (v \lor C_1) \land \cdots \land (v \lor C_m) \land (\lnot v \lor x_1) \land \cdots \land (\lnot v \lor x_n),$$ where $v$ is a new variable and $x_1,\ldots,x_n$ are the original variables. Then $\psi$ has a unique satisfying assignment (in which all variables are true) iff $\phi$ is unsatisfiable.
As for your first question, if $\mathsf{P} = \mathsf{NP}$ then $\mathsf{P} = \mathsf{UP} = \mathsf{US}$, so no unconditional separation is currently known. If you're interested in conditional separations, I suggest asking a new question focused just on that issue.