Extending the problem, admit the possibility of equality of cardinals, so consider that $\kappa \leq \mu$. If indeed $\kappa=\mu$, then the collection of subsets with one member, namely $\mu$, suffices. But that is the only case.
For we borrow the argument from the comments. Let $U$ be a proper subset of $\mu$ that is part of a suitable collection, and let $V$ be $\kappa$-sized and contained in $U$. If $\kappa$ is infinite, pick some element of $\mu \setminus U$ and consider a set $U'$ from the collection that contains the $\kappa$-sized set which is $V$ union (the singleton set containing) this element. Then both $U$ and $U'$ contain $V$, which means condition two does not hold, and so a suitable collection cannot have a proper subset of $\mu$.
For finite $\kappa$ and infinite $\lambda \lt \mu$, here is an idea which may provide a suitable collection, but some work needs to be done. Namely, well order all of the $\lambda$-sized subsets of $\mu$, and pick the least allowed $\lambda$ sized subset $U$, and then "throw out" the $\lambda$ sets that intersect $U$ in a subset of size at least $\kappa$. Conditions 1) and 2) are partly satisfied during the construction, but it remains to show that every $\kappa$ subset is uniquely covered. Since $\mu$ is strictly larger than $\lambda$, this should be possible to show.
Gerhard "Infinitely Elementary, My Dear Dominic" Paseman, 2018.12.20.