Exercise on Spectrum of Commutative Elements in a Banach Algebra 
Given an element $A$ of a Banach algebra $V$ and $\epsilon > 0$, prove that if $0 \in \textrm{sp}(A)$, then there is $\delta > 0$ such that if $B \in V$ commutes with $A$ and $||A-B|| < \delta$, then there is a $\lambda$ in $\textrm{sp}(B)$ with $|\lambda | < \epsilon$.

My Attempt
My goal was to prove that by making $\delta$ small enough, we can guarantee $r(B)$ (spectral radius of $B$) to be less than $\epsilon$, and then use the fact that the spectrum is non-empty to prove the existence of such a $\lambda$.
Since $A$ and $B$ commutes, then $B-A$ and $A$ commutes, so we have:
$$r(B) = r(B-A+A) \leq r(B-A) + r(A) \leq ||B-A|| + r(A).$$
I am not sure how to proceed from here. I do not think this approach will work since I think the RHS of the above expression will not always be less than $\epsilon$ since $r(A)$ can be anything. However, I cannot think of any other approaches to take.
Could someone give a hint on how to approach this question?
 A: Lemma.  Given a unital Banach algebra $V$, and a
commutative subalgebra $W$,  there exists another commutative
subalgebra $W_1$, containing $W$, such that for any element $a$ in $W_1$, one has that $a$ is invertible relative to $W_1$
iff $a$ is invertible relative to $V$.   In particular, $$ \text{sp}_{V}(a)=\text{sp}_{W_1}(a),  $$
for every $a$ in $W_1$.
Proof. For every subset $S\subseteq V$, define the commutant of $S$ by
$$
  S'=\{a\in  V: as=sa: \text{ for all } s\in S\}.
  $$
It is
easy to see that
(1) $S'$ is always a unital subalgebra,
(2) $S$ is commutative iff $S\subseteq S'$,
(3) if $S\subseteq T$ then $S'\supseteq T'$.
(4) if $a\in S'$, and $a$ is invertible,  then $a^{-1}\in S'$.
Now, given $W$ as in the statement, we claim that  $W_1:= W''$ (that is, the commutant of the commutant of $W$) satisfies all of the required conditions.
First  observe that $W\subseteq W''$ by the following very trivial (if clumsy) reason:  every element of $W$ commutes with
everything that commutes with the elements of $W$.
Since $W$ is commutative, we deduce from (2) that  $W\subseteq W'$.  Using
(3) we get $W'\supseteq W''$, and   using (3) again we get $W''\subseteq W'''$. So the converse part of (2) implies that $W''$ is commutative.
Finally the last condition in the statement regarding invertible elements follows immediately from (4). QED
Back to the original question,  consider the commutative Banach algebra $W$ generated by $A$  and $B$,  and let $W_1$ be
as in the Lemma.  Then, for every element $a$ in $W_1$, we have that
$$
  \text{sp}_{V}(a)=\text{sp}_{W_1}(a) = \{\phi(a): \phi\in \text{Hom}(W_1, \mathbb C)\},
  $$
so the end result follows easily by the continuity of complex homomorphisms.
