$X$ and $U$ are Banach spaces. A linear map $\textbf{C} : X \rightarrow U$ is called compact if the image $\textbf{C}B$ of the unit ball $B$ in X is precompact in $U$. A subset S of a complete metric space is called precompact if its closure is compact
I cannot come up with an counter to prove that the strong limit of a sequence of compact operators need not to be compact.
Since all closed subsets of a compact space are closed, every linear map to a compact space is compact. Hence $U$ should not be compact at least..