Let $l_{k}^{2}$ denote the k-dimensional Hilbert space and $\oplus_{1}^{\infty} l_{k}^{2}$ be the infinite direct sum of $l_{k}^{2}$. Let $P_{M}\in B(l_{k}^{2}\oplus[\bigoplus_{n=1}^{\infty}l_{k}^{2}])$ be the orthogonal projection from $l_{k}^{2}\oplus[\bigoplus_{n=1}^{\infty}l_{k}^{2}]$ onto the subspace $$l_{k}^{2}\oplus[\bigoplus_{n=1}^{M}l_{k}^{2}],$$ if there exists a unitary $U\in\mathbb{C}1_{l_{k}^{2}\oplus[\bigoplus_{n=1}^{\infty}l_{k}^{2}]}+K(l_{k}^{2}\oplus[\bigoplus_{n=1}^{\infty}l_{k}^{2}])$ (The $K(l_{k}^{2}\oplus[\bigoplus_{n=1}^{\infty}l_{k}^{2}]$ denotes the compact operators on $l_{k}^{2}\oplus[\bigoplus_{n=1}^{\infty}l_{k}^{2}]$), then
Question 1 Can we verify that $||P_{M}U-UP_{M}||\rightarrow 0$ as $M\rightarrow \infty$?
Question 2 If $||P_{M}U-UP_{M}||\rightarrow 0$ is true, can we find unitaries $U_{M}\in B(l_{k}^{2}\oplus[\bigoplus_{n=1}^{M}l_{k}^{2}])$ such that $||U_{M}-P_{M}UP_{M}||\rightarrow 0$ as $M\rightarrow 0$? (By standard perturbation theory?)
Remark: This two question comes from a book "C*-algebras and Finite-Dimensional Approximations" P266, the proof of Theorem 8.1.8. The notations are a little bit complicated, but the question may not be use so much conditions.
The situation is the following: you have $U=\lambda I+T$, with $T$ compact. And $\{P_M\}$ is a sequence of finite-rank projections such that $P_M\nearrow I$.
So, given $\varepsilon>0$, by the compactness of $T$ you can write $T=P_MTP_M+T_0$, with $\|T_0\|<\varepsilon$. Then $$ \|P_MU-UP_M\|=\|P_MT_0-T_0P_M\|<2\varepsilon. $$
For your second question, $P_MUP_M$ is a finite-rank operator. We can see it as an element in the algebra $P_M\,B(H)\,P_M$, which is isomorphic to $M_n(\mathbb C)$ for appropriate $n$. Let $P_MUP_M=U_M|P_MUP_M|$ be the polar decomposition; we can always take $U_M$ to be a unitary because we are dealing with matrices here. We have \begin{align} \|U_M-P_MUP_M\|&=\|U_MP_M-P_MUP_M\|=\|U_MP_M-U_M|P_MUP_M|\|\\ &=\|P_M-|P_MUP_M|\|\leq\|P_M-|P_MUP_M|^2\|\\ &=\|P_M-P_MU^*P_MUP_M\|=\|P_M(P_M-U^*P_MU)P_M\|\\ &\leq\|P_M-U^*P_MU\|=\|UP_M-P_MU\|. \end{align} For the first inequality above, note that in the matrix algebra $P_M$ is the identity, so the inequality is, via functional calculus, just the fact that $1-|t|\leq1-|t|^2$ when $|t|\leq1$.