[proofplan]
Set $A:=N-\lambda I_H$ and identify the spectrum of $A^{-1}$ using the [rational spectral mapping theorem](/theorems/8392) applied to the rational function $z\mapsto (z-\lambda)^{-1}$. Since $A^{-1}$ is again normal, the norm equality for normal operators converts the operator norm of $A^{-1}$ into the maximum modulus of its spectrum. The remaining step is a compactness computation: because $\lambda\notin\sigma(N)$ and $\sigma(N)$ is compact and nonempty, the minimum of $|z-\lambda|$ on $\sigma(N)$ is exactly $\operatorname{dist}(\lambda,\sigma(N))$.
[/proofplan]
[step:Show that the inverse resolvent is a normal operator]
Define
\begin{align*}
A:=N-\lambda I_H\in\mathcal L(H).
\end{align*}
Since $\lambda\in\rho(N)$, the operator $A$ is invertible and $A^{-1}=(N-\lambda I_H)^{-1}$ is well-defined.
We first verify that $A$ is normal. Since $N$ is normal, $N^*N=NN^*$. Also
\begin{align*}
A^*=N^*-\overline{\lambda}I_H.
\end{align*}
Expanding both products gives
\begin{align*}
A^*A=(N^*-\overline{\lambda}I_H)(N-\lambda I_H)
=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H
\end{align*}
and
\begin{align*}
AA^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H)
=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H.
\end{align*}
These are equal because $N^*N=NN^*$. Thus $A$ is normal.
Since $A$ is invertible and normal, $A^{-1}$ is normal. Indeed, from $A^*A=AA^*$ and invertibility of $A$ and $A^*$, taking inverses gives
\begin{align*}
(A^*A)^{-1}=(AA^*)^{-1}.
\end{align*}
Using the inverse of a product, this is
\begin{align*}
A^{-1}(A^*)^{-1}=(A^*)^{-1}A^{-1}.
\end{align*}
Since $(A^{-1})^*=(A^*)^{-1}$, this says
\begin{align*}
A^{-1}(A^{-1})^*=(A^{-1})^*A^{-1}.
\end{align*}
Hence $A^{-1}$ is normal.
[guided]
The first point is that the norm formula we want to use later applies to normal operators, so we must check that the resolvent operator itself is normal. Define
\begin{align*}
A:=N-\lambda I_H\in\mathcal L(H).
\end{align*}
Because $\lambda\in\rho(N)$, by the definition of the resolvent set the operator $N-\lambda I_H$ is invertible. Thus $A^{-1}=(N-\lambda I_H)^{-1}$ exists as a bounded operator on $H$.
We check normality directly. The adjoint of $A$ is
\begin{align*}
A^*=N^*-\overline{\lambda}I_H,
\end{align*}
using conjugate-linearity of the Hilbert-space adjoint in scalars. Therefore
\begin{align*}
A^*A=(N^*-\overline{\lambda}I_H)(N-\lambda I_H)
=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H.
\end{align*}
Similarly,
\begin{align*}
AA^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H)
=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H.
\end{align*}
The only possible difference between these two expressions is the pair $N^*N$ and $NN^*$. Since $N$ is normal, $N^*N=NN^*$. Hence $A^*A=AA^*$, so $A$ is normal.
Now we pass normality through inversion. Since $A$ is invertible, $A^*$ is invertible and $(A^{-1})^*=(A^*)^{-1}$. Taking inverses in the equality $A^*A=AA^*$ gives
\begin{align*}
(A^*A)^{-1}=(AA^*)^{-1}.
\end{align*}
The inverse-of-a-product rule gives
\begin{align*}
A^{-1}(A^*)^{-1}=(A^*)^{-1}A^{-1}.
\end{align*}
Replacing $(A^*)^{-1}$ by $(A^{-1})^*$, we obtain
\begin{align*}
A^{-1}(A^{-1})^*=(A^{-1})^*A^{-1}.
\end{align*}
This is exactly the normality condition for $A^{-1}$.
[/guided]
[/step]
[step:Compute the spectrum of the inverse resolvent]
Define the rational function
\begin{align*}
r:\mathbb C\setminus\{\lambda\}&\to\mathbb C
\end{align*}
\begin{align*}
z&\mapsto \frac{1}{z-\lambda}.
\end{align*}
Because $\lambda\in\rho(N)$, we have $\lambda\notin\sigma(N)$, so the denominator $z-\lambda$ is nonzero for every $z\in\sigma(N)$. Thus the hypotheses of the [Rational Spectral Mapping Theorem][citetheorem:8392] apply to $N$ and $r$. Hence
\begin{align*}
\sigma(A^{-1})=\sigma((N-\lambda I_H)^{-1})=r(\sigma(N))
=\left\{\frac{1}{z-\lambda}:z\in\sigma(N)\right\}.
\end{align*}
[/step]
[step:Apply the norm equality for normal operators to the inverse resolvent]
Since $A^{-1}$ is normal, the [Norm Equality for Normal Operators][citetheorem:8418] applies to $A^{-1}$. Therefore
\begin{align*}
\|A^{-1}\|_{\mathcal L(H)}
=\max_{\mu\in\sigma(A^{-1})}|\mu|.
\end{align*}
Using the spectral computation from the previous step, this becomes
\begin{align*}
\|A^{-1}\|_{\mathcal L(H)}
=\max_{z\in\sigma(N)}\left|\frac{1}{z-\lambda}\right|.
\end{align*}
Since $\lambda\notin\sigma(N)$, every denominator is nonzero, so
\begin{align*}
\|A^{-1}\|_{\mathcal L(H)}
=\max_{z\in\sigma(N)}\frac{1}{|z-\lambda|}.
\end{align*}
[/step]
[step:Rewrite the maximum as the reciprocal of the spectral distance]
Since $H$ is nonzero and $N\in\mathcal L(H)$, the spectrum $\sigma(N)$ is nonempty and compact. The map
\begin{align*}
g:\sigma(N)&\to\mathbb R
\end{align*}
\begin{align*}
z&\mapsto |z-\lambda|
\end{align*}
is continuous. Because $\sigma(N)$ is compact, $g$ attains its minimum at some point $z_0\in\sigma(N)$. Since $\lambda\notin\sigma(N)$, this minimum is positive:
\begin{align*}
g(z_0)=\min_{z\in\sigma(N)}|z-\lambda|>0.
\end{align*}
By definition of distance from a point to a set,
\begin{align*}
\operatorname{dist}(\lambda,\sigma(N))
=\inf_{z\in\sigma(N)}|\lambda-z|
=\min_{z\in\sigma(N)}|z-\lambda|.
\end{align*}
The function $t\mapsto 1/t$ is decreasing on $(0,\infty)$, so
\begin{align*}
\max_{z\in\sigma(N)}\frac{1}{|z-\lambda|}
=\frac{1}{\min_{z\in\sigma(N)}|z-\lambda|}
=\frac{1}{\operatorname{dist}(\lambda,\sigma(N))}.
\end{align*}
Since $A^{-1}=(N-\lambda I_H)^{-1}$, we conclude
\begin{align*}
\|(N-\lambda I_H)^{-1}\|_{\mathcal L(H)}
=\frac{1}{\operatorname{dist}(\lambda,\sigma(N))}.
\end{align*}
[/step]
