Being Close to a Unitary in a C$^{*}$-algebra Implies Being Close to a Unitary in a Sub-C$^{*}$-Algebra

My question is Exercise 2.8. from Rordam's blue book. It is the following:

Suppose $A$ is a unital C$^{*}$-algebra and $\epsilon>0$. Find a $\delta>0$ such that if $B$ is a sub-C$^{*}$-algebra of $A$ containing $1_{A}$ and $\|u-b\|\leq \delta$ for some $b\in B$ and unitary $u\in A$, then there is a unitary $v\in B$ satisfying $\|u-v\|\leq\epsilon$.

This is the second half of the exercise. I was able to solve the first part and there is also a solution here: almost unitaries are close to a unitary element

My attempt at the second part: Choosing $\delta<1$, we obtain $$\|1_{A}-bu^{*}\|=\|(u-b)u^{*}\|\leq\|u-b\|\|u^{*}\|=\|u-b\|<1,$$ so that $bu^{*}$ is invertible in $A$. Similarly, I showed that $ub^{*}, b^{*}u$, and $u^{*}b$ are also invertible in $A$. But this gives that $b^{*}b$ and $bb^{*}$ are invertible in $A$ and, consequently, $b$ is invertible in $B$. My reason for doing this was to use the polar decomposition of $b$ to write $b=v|b|$ for some unitary $v\in B$. In hopes of getting $\|u-v\|\leq \epsilon$, I estimated:

$$\|u-v\|\leq\|u-b\|+\|b-v\|\leq \delta+\|b-v\|.$$

Using the spectral mapping theorem, I deduced that $$\|b-v\|=\|v|b|-v\|\leq\||b|-1_{A}\|\leq\max\{\delta,1\},$$ which won't do it since I have the nasty $1$ there.

If anyone has any idea how to solve this problem, either with my approach or something different, I'd be very grateful.

You have $$1_A\leq 1_A+|b|.$$ So $1_A+|b|$ is invertible, and $\|(1_A+|b|)^{-1}\|\leq1$.
Then  And \begin{align} \|\,|b|-1_A\| &=\|\,(|b|^2-1_A)(|b|+1_A)^{-1}\|\leq\|\,|b|^2-1_A\|\\ \ \\ &=\|b^*b-u^*u\|=\|b^*b-u^*b+u^*b-u^*u\|\\ \ \\ &\leq\|b\|\,\|b-u\|+\|b-u\|. \end{align}