Let $H$ be an infinite dimensional Hilbertspace and $U(H)$ the group of unitaries endowed with the norm topology.
Is $U(H)$ connected?
The following is a generalisation of the proof in the finite-dimensional case but will not work for an arbitrary unitary:
If we consider a unitary $u$, then $\Bbb1, u, u^*$ generate an abelian $C^*$-algebra isomorphic to $C(\sigma(u))$, here $\sigma(u)\subset S^1$ and $u$ corresponds to the function $\sigma(u)\to\Bbb C$, $z\mapsto z$. In the event that $\sigma(u)\neq S^1$ one can define a homotopy on $S^1$ that deforms $\sigma(u)$ to a point. This homotopy will correspond to a path that brings $u$ to $\Bbb1$ via unitaries in $C(\sigma(u))$ and thus in $U(H)$.
For $\sigma(u)=S^1$ such a homotopy is impossible, and $u$ is not connected to $\Bbb1$ in $C(\sigma(u))$. However if there exists a path in $U(H)$ connecting $u$ to a unitary with spectrum smaller than $S^1$, the above will give the result.
Another approach I had thought about was looking at the Cayley transform, however this does not have image all of $U(H)$. Finding a logarithm (ie a self-adjoint $A$ with $\exp(A)=u$) would give a path from $u$ to $\Bbb 1$ via $\exp(tA)$, but this path is only strongly continuous and not norm continuous.
$B(H)$ is a von Neumann algebra and you are allowed to use Borel functional calculus, so every unitary is connected to the identity.