If you consider the multiplication by a continuous function as a subset of $B(L^2(\mathbb{T}))$, is this norm closed in operator norm? i.e. if B is an operator in $B(L^2(\mathbb{T}))$ with $||B-M_{\phi_n}||\rightarrow 0$ for $\phi_n$ continuous functions then is B also a multiplication by a continuous function?
I have the impression that this is not true but it seems to be assumed to show that the reduced group C* algebra of $\mathbb{Z}$ is $C(\mathbb{T})$.
Yes, $\{M_\phi\mid \phi\in C(\mathbb T)\}$ is a closed subalgebra of $B(L^2(\mathbb T))$. One way to see this is by showing that the map $\phi\mapsto M_\phi$ is an isometry, hence its image is complete, and thus closed.