The following question is from Exercise $\S 10.11$ in A Course in Operator Theory written by John B. Conway:
Show that, given a normal operator in $B(H)$, there is a sequence of invertible normal operators $N_n$ such that $\|N_n-N\|\rightarrow 0$ and the sequence can be chosen such that, for each $k\in\mathbb{N}$, $\sigma(N_k)$ is either: $1)$: finite; $2)$: a set bounded by finitely many smooth curves; $3)$: a finite collection of line segments in the complex plane. Further, can each $\sigma(N_k)$ be connected? What if $\sigma(N)$ is connected?
If we let $E$ to be the spectral measure defined on $\sigma(N)$, then I can write: $$ N = \int_{\sigma(N)}z\,dE(z) $$
and hence $N$ can be always approximated (uniformly) by a sequence of simple functions, which answers part 1), or $f_k(N)$ where $f_k$ is defined by:
$$ f_k(z) = \begin{cases}\frac{1}{k},\hspace{0.5cm}\vert\,z\,\vert\leq\frac{1}{k}\\ z,\hspace{0.62cm}\vert\,z\,\vert > \frac{1}{k} \end{cases} $$
Below are my questions:
i): In general, how to solve part 2) and 3)? Especially in part 3), by "line segments", I suppose it means "a segment of a straight line" but not just a general curve. When $\sigma(N)$ is connected, $f_k(N)$ defined above can answer part 2).
ii): Beside approximating $N$ by $f_k(N)$, are there other ways to approximating $N$ by a sequence of invertible normal operators such that each $\sigma(N_k)$ is connected?
iii): part 2) does not require that $\sigma(N)$ has at most countably components. What can we tell about the point (resp. continuous) spectrum when $\sigma(N)$ has uncountably components?
Given $k\in\mathbb N$, there exists $f_k:\sigma(N)\to\mathbb C$, simple, with $\|f_k-\operatorname{id}\|<\frac1k$. We can write $$ f_k=\sum_{j=1}^{m_k}\alpha_{k,j}\,1_{E_{k,j}}, $$ where $\bigcup_jE_{k,j}=\sigma(N)$, and $\alpha_{k,j}\in\sigma(N)$ for all $k,j$. If $\alpha_{k,j}=0$ for any $j$, we replace it with $1/k$. With this assumption we get $\|f_k-\operatorname{id}\|<\frac2k$ and $|f_k|\geq\frac1k$.
So $N_k=f_k(N)$ satisfies $\|N_k-N\|=\|f_k-\operatorname{id}\|<\frac2k$ for all $k$. And $$ \sigma(N_k)=\operatorname{ran}f_k=\{\alpha_{k,1},\ldots,\alpha_{k,m_k}\}. $$ That is, it is always possible to get $\sigma(N_k)$ finite.
In general, one cannot expect $\sigma(N_k)$ to be connected. For instance let $P$ be a projection and let $N=P+2(I-P)$. Then $\sigma(N)=\{1,2\}$. Suppose that $M$ is normal, invertible, and $\|M-N\|<\frac1{10}$. Then, if $|1-\lambda|\geq\frac1{10}$ and $|2-\lambda|\geq\frac1{10}$,
$$ \|(M-\lambda)-(N-\lambda)\|=\|M-N\|<\frac1{10}\leq\max\{|1-\lambda|,|2-\lambda|\}=\|(N-\lambda)^{-1}\|^{-1} $$ and then $M-\lambda$ is invertible. This shows that $$\sigma(M)\subset B_{1/10}(1)\cup B_{1/10}(2).$$ And this prevents $\sigma(M)$ from being connected, because it has to have points in both balls. Indeed, if for instance we had $\sigma(M)\subset B_{1/10}(1)$, then $\|I-M\|<\frac1{10}$. Then, since $$ \|N-I\|=\|I-P\|=1, $$ we have $$ \|N-M\|\geq\|N-I\|-\|I-M\|\geq\|N-I\|-\frac1{10}=1-\frac1{10}. $$ A similar computation leads to a contradiction if $\sigma(M)\subset B_{1/10}(2)$.
When $\sigma(N)$ is connected, one can get $\sigma(N_k)$ connected for all $k$. The function you defined may not work because $1/k$ may be isolated, but one can tweak the function so that its range on $\sigma(N)$ is connected.
As for your last question, for a normal operator the residual spectrum is always empty.