Can $U(n)$ be constructed as a countable union of proper Lie subgroups of $U(n)$? Can $U(n)$ be constructed as a countable union of proper Lie subgroups of $U(n)$ for any $n$?
As I see it, the obvious answer is "no", since the dimension of the Lie algebra for any proper subgroup of $U(n)$ must be strictly less than $\dim\mathfrak{u}(n)$, meaning that the dimension of the subgroup must be strictly less than the dimension of $U(n)$, and a smooth manifold of dimension $d$ cannot be covered by a countable union of submanifolds of dimension $k<d$.
Is this intuition correct, or am I missing something critical?
(Also, if any of my terminology here is in error or seems awkward, I'd appreciate a heads up. With thanks.)

Your intuition is correct. More generally, no differentiable manifold can be written as a countable union of closed submanifolds of smaller dimension. This, together with the fact that $U(n)$ is connected (which imples that no subgroup of $U(n)$ has the same dimension of $\dim U(n)$), is enough to prove what you want to prove.
