[proofplan]
Fix a point $\lambda_0 \in \rho(T)$ and factor $T-\lambda I_X$ through the already invertible operator $T-\lambda_0 I_X$. The remaining factor is $I_X-(\lambda-\lambda_0)R(\lambda_0,T)$, which is invertible whenever its perturbation has operator norm strictly less than $1$; this is proved directly by the Neumann series. The same factorization gives a locally convergent [power series](/page/Power%20Series) for $R(\lambda,T)$ in the operator norm, and this local power-series representation is exactly holomorphy of the resolvent.
[/proofplan]
[step:Handle the degenerate zero-space case separately]
If $X=\{0\}$, then $\mathcal{L}(X)$ consists of the single operator, which is also the identity operator on $X$. Hence $T-\lambda I_X$ is invertible for every $\lambda \in \mathbb{C}$, with inverse equal to the unique operator on $X$. Therefore $\rho(T)=\mathbb{C}$, which is open, and $R_T$ is the constant map from $\mathbb{C}$ to the zero [Banach space](/page/Banach%20Space) $\mathcal{L}(X)$, hence holomorphic in the operator-norm topology.
For the rest of the proof, assume $X\ne \{0\}$.
[/step]
[step:Factor nearby shifted operators through a fixed resolvent point]
Fix $\lambda_0 \in \rho(T)$, and define
\begin{align*}
A_0 := T-\lambda_0 I_X \in \mathcal{L}(X),
\end{align*}
so that $A_0$ is invertible in $\mathcal{L}(X)$ and
\begin{align*}
R_0 := R(\lambda_0,T)=A_0^{-1}\in \mathcal{L}(X).
\end{align*}
For each $\lambda \in \mathbb{C}$, define the scalar
\begin{align*}
z := \lambda-\lambda_0 \in \mathbb{C}.
\end{align*}
Then
\begin{align*}
T-\lambda I_X = T-\lambda_0 I_X-(\lambda-\lambda_0)I_X.
\end{align*}
Using $A_0R_0=I_X$, this becomes
\begin{align*}
T-\lambda I_X=A_0-zA_0R_0=A_0(I_X-zR_0).
\end{align*}
Thus the invertibility of $T-\lambda I_X$ reduces to the invertibility of $I_X-zR_0$.
[/step]
[step:Invert the small perturbation by a Neumann series]
Assume
\begin{align*}
|z|\|R_0\|_{\mathcal{L}(X)}<1.
\end{align*}
For each integer $n\ge 0$, define
\begin{align*}
S_n := \sum_{k=0}^{n} z^k R_0^k \in \mathcal{L}(X).
\end{align*}
Since $\mathcal{L}(X)$ is a Banach algebra under the operator norm and
\begin{align*}
\|z^kR_0^k\|_{\mathcal{L}(X)}\le |z|^k\|R_0\|_{\mathcal{L}(X)}^k,
\end{align*}
the numerical geometric series
\begin{align*}
\sum_{k=0}^{\infty}|z|^k\|R_0\|_{\mathcal{L}(X)}^k
\end{align*}
converges. Hence the operator series defining
\begin{align*}
S := \sum_{k=0}^{\infty} z^k R_0^k \in \mathcal{L}(X)
\end{align*}
converges absolutely in the operator norm.
For every $n\ge 0$, multiplication in $\mathcal{L}(X)$ gives
\begin{align*}
(I_X-zR_0)S_n = I_X-z^{n+1}R_0^{n+1}.
\end{align*}
Similarly,
\begin{align*}
S_n(I_X-zR_0)=I_X-z^{n+1}R_0^{n+1}.
\end{align*}
Because
\begin{align*}
\|z^{n+1}R_0^{n+1}\|_{\mathcal{L}(X)}\le |z|^{n+1}\|R_0\|_{\mathcal{L}(X)}^{n+1}\to 0,
\end{align*}
passing to the operator-norm limit yields
\begin{align*}
(I_X-zR_0)S=I_X
\end{align*}
and
\begin{align*}
S(I_X-zR_0)=I_X.
\end{align*}
Therefore $I_X-zR_0$ is invertible in $\mathcal{L}(X)$ with inverse $S$.
[guided]
The goal is to prove that the factor $I_X-zR_0$ is invertible when $z$ is small. Since $R_0$ is bounded, the operator $zR_0$ has norm
\begin{align*}
\|zR_0\|_{\mathcal{L}(X)}=|z|\|R_0\|_{\mathcal{L}(X)}.
\end{align*}
The hypothesis says this norm is strictly less than $1$, so we try to invert $I_X-zR_0$ by the same formula used for scalar geometric series.
For each integer $n\ge 0$, define the partial sum
\begin{align*}
S_n := \sum_{k=0}^{n} z^k R_0^k \in \mathcal{L}(X).
\end{align*}
This is a well-defined bounded operator because it is a finite sum of bounded operators. To pass from partial sums to an infinite operator series, we need convergence in the operator norm. The Banach algebra estimate gives
\begin{align*}
\|z^kR_0^k\|_{\mathcal{L}(X)}\le |z|^k\|R_0\|_{\mathcal{L}(X)}^k.
\end{align*}
The right-hand side is the $k$th term of a geometric series with ratio $|z|\|R_0\|_{\mathcal{L}(X)}<1$. Therefore the series
\begin{align*}
\sum_{k=0}^{\infty} z^kR_0^k
\end{align*}
converges absolutely in the operator norm. Since $\mathcal{L}(X)$ is complete, its sum is an operator
\begin{align*}
S := \sum_{k=0}^{\infty} z^k R_0^k \in \mathcal{L}(X).
\end{align*}
Now we verify that $S$ is really the inverse. Multiplying the partial sum by $I_X-zR_0$ causes all intermediate terms to cancel:
\begin{align*}
(I_X-zR_0)S_n = I_X-z^{n+1}R_0^{n+1}.
\end{align*}
The same cancellation on the other side gives
\begin{align*}
S_n(I_X-zR_0)=I_X-z^{n+1}R_0^{n+1}.
\end{align*}
The remainder term tends to zero in the operator norm because
\begin{align*}
\|z^{n+1}R_0^{n+1}\|_{\mathcal{L}(X)}\le |z|^{n+1}\|R_0\|_{\mathcal{L}(X)}^{n+1}\to 0.
\end{align*}
Taking the operator-norm limit in the two identities gives
\begin{align*}
(I_X-zR_0)S=I_X
\end{align*}
and
\begin{align*}
S(I_X-zR_0)=I_X.
\end{align*}
Thus $S$ is both a left and right inverse for $I_X-zR_0$, so $I_X-zR_0$ is invertible in $\mathcal{L}(X)$.
[/guided]
[/step]
[step:Deduce that the resolvent set is open]
Since $X\ne\{0\}$ and $A_0$ is invertible, $R_0=A_0^{-1}$ is nonzero. Hence
\begin{align*}
r_0 := \frac{1}{\|R_0\|_{\mathcal{L}(X)}}>0
\end{align*}
is well-defined. Define the open ball
\begin{align*}
B(\lambda_0,r_0) := \{\lambda \in \mathbb{C} : |\lambda-\lambda_0| < r_0\}.
\end{align*}
If $\lambda \in B(\lambda_0,r_0)$, then $z=\lambda-\lambda_0$ satisfies
\begin{align*}
|z|\|R_0\|_{\mathcal{L}(X)}<1.
\end{align*}
By the previous step, $I_X-zR_0$ is invertible in $\mathcal{L}(X)$. Since $A_0$ is invertible and
\begin{align*}
T-\lambda I_X=A_0(I_X-zR_0),
\end{align*}
the product $T-\lambda I_X$ is invertible in $\mathcal{L}(X)$. Therefore $\lambda\in\rho(T)$.
We have shown that for every $\lambda_0\in\rho(T)$ there exists $r_0>0$ such that
\begin{align*}
B(\lambda_0,r_0)\subset \rho(T).
\end{align*}
Thus $\rho(T)$ is open in $\mathbb{C}$.
[/step]
[step:Expand the resolvent as a locally convergent operator power series]
For $\lambda \in B(\lambda_0,r_0)$, the inverse of $T-\lambda I_X=A_0(I_X-zR_0)$ is
\begin{align*}
R(\lambda,T)=(I_X-zR_0)^{-1}A_0^{-1}.
\end{align*}
Using $A_0^{-1}=R_0$ and the Neumann series from the previous step gives
\begin{align*}
R(\lambda,T)=\left(\sum_{k=0}^{\infty}z^kR_0^k\right)R_0.
\end{align*}
Since the series converges absolutely in the operator norm, multiplication on the right by $R_0$ is continuous in the operator norm. Therefore
\begin{align*}
R(\lambda,T)=\sum_{k=0}^{\infty}z^kR_0^{k+1}.
\end{align*}
Recalling that $z=\lambda-\lambda_0$ and $R_0=R(\lambda_0,T)$, this is
\begin{align*}
R(\lambda,T)=\sum_{k=0}^{\infty}(\lambda-\lambda_0)^kR(\lambda_0,T)^{k+1}.
\end{align*}
The series converges in the operator norm for every $\lambda$ satisfying
\begin{align*}
|\lambda-\lambda_0|<\frac{1}{\|R(\lambda_0,T)\|_{\mathcal{L}(X)}}.
\end{align*}
[/step]
[step:Conclude holomorphy in the operator norm]
The preceding step shows that at each point $\lambda_0\in\rho(T)$, the map $R_T:\rho(T)\to\mathcal{L}(X)$ agrees on a neighbourhood of $\lambda_0$ with an operator-norm convergent power series in the complex variable $\lambda-\lambda_0$, namely
\begin{align*}
R_T(\lambda)=\sum_{k=0}^{\infty}(\lambda-\lambda_0)^kR(\lambda_0,T)^{k+1}.
\end{align*}
By the definition of a holomorphic Banach-space-valued function as a function locally representable by a norm-convergent complex power series, $R_T$ is holomorphic on $\rho(T)$ in the operator-norm topology. This completes the proof.
[/step]
