[proofplan]
We fix $\lambda_0\in\rho(T)$ and prove that every complex number sufficiently close to $\lambda_0$ also belongs to the resolvent set. The key point is to factor $T-\lambda I$ into the already invertible operator $T-\lambda_0 I$ and a small perturbation of the identity. The small factor is inverted by an explicit Neumann series, whose convergence follows from the [Banach space](/page/Banach%20Space) completeness of $\mathcal{L}(X)$.
[/proofplan]
[step:Factor $T-\lambda I$ through the known inverse at $\lambda_0$]
Since $\lambda_0\in\rho(T)$, the [bounded linear operator](/page/Bounded%20Linear%20Operator)
\begin{align*}
T-\lambda_0 I:X\to X
\end{align*}
is invertible, with inverse $R(\lambda_0,T)\in\mathcal{L}(X)$. Let $\lambda\in\mathbb C$ satisfy
\begin{align*}
|\lambda-\lambda_0|<\frac{1}{\|R(\lambda_0,T)\|_{\mathcal{L}(X)}}.
\end{align*}
Define
\begin{align*}
A=(\lambda-\lambda_0)R(\lambda_0,T)\in\mathcal{L}(X).
\end{align*}
Then, using $(T-\lambda_0 I)R(\lambda_0,T)=I$, we compute in $\mathcal{L}(X)$:
\begin{align*}
(T-\lambda_0 I)(I-A)=T-\lambda_0 I-(\lambda-\lambda_0)(T-\lambda_0 I)R(\lambda_0,T).
\end{align*}
Therefore
\begin{align*}
(T-\lambda_0 I)(I-A)=T-\lambda_0 I-(\lambda-\lambda_0)I=T-\lambda I.
\end{align*}
[/step]
[step:Invert the small perturbation of the identity by a Neumann series]
The operator $A$ satisfies
\begin{align*}
\|A\|_{\mathcal{L}(X)}=|\lambda-\lambda_0|\,\|R(\lambda_0,T)\|_{\mathcal{L}(X)}<1.
\end{align*}
For each $n\in\mathbb N\cup\{0\}$, let $A^n$ denote the $n$-fold product of $A$, with $A^0=I$. Since $\mathcal{L}(X)$ is a Banach space and
\begin{align*}
\sum_{n=0}^{\infty}\|A^n\|_{\mathcal{L}(X)}\leq \sum_{n=0}^{\infty}\|A\|_{\mathcal{L}(X)}^n<\infty,
\end{align*}
the series
\begin{align*}
S=\sum_{n=0}^{\infty}A^n
\end{align*}
converges in $\mathcal{L}(X)$ to some operator $S\in\mathcal{L}(X)$.
For each $N\in\mathbb N\cup\{0\}$, define the partial sum
\begin{align*}
S_N=\sum_{n=0}^{N}A^n.
\end{align*}
Then
\begin{align*}
(I-A)S_N=I-A^{N+1}
\end{align*}
and
\begin{align*}
S_N(I-A)=I-A^{N+1}.
\end{align*}
Because $\|A^{N+1}\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^{N+1}\to 0$, passing to the limit in operator norm gives
\begin{align*}
(I-A)S=I
\end{align*}
and
\begin{align*}
S(I-A)=I.
\end{align*}
Thus $I-A$ is invertible in $\mathcal{L}(X)$, with inverse $S$.
[guided]
We need to prove that $I-A$ is invertible, where
\begin{align*}
A=(\lambda-\lambda_0)R(\lambda_0,T).
\end{align*}
The reason this is possible is that $A$ is small in operator norm. Indeed, by homogeneity of the operator norm,
\begin{align*}
\|A\|_{\mathcal{L}(X)}=|\lambda-\lambda_0|\,\|R(\lambda_0,T)\|_{\mathcal{L}(X)}<1.
\end{align*}
The candidate inverse is the Neumann series. For each $n\in\mathbb N\cup\{0\}$, let $A^n$ denote the $n$-fold product of $A$, with $A^0=I$. Since operator norm is submultiplicative, we have
\begin{align*}
\|A^n\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^n.
\end{align*}
The scalar geometric series $\sum_{n=0}^{\infty}\|A\|_{\mathcal{L}(X)}^n$ converges because $\|A\|_{\mathcal{L}(X)}<1$. Hence
\begin{align*}
\sum_{n=0}^{\infty}\|A^n\|_{\mathcal{L}(X)}<\infty.
\end{align*}
Since $X$ is Banach, the operator space $\mathcal{L}(X)$ is also Banach in the operator norm, so absolute convergence of this series implies convergence in $\mathcal{L}(X)$. Define
\begin{align*}
S=\sum_{n=0}^{\infty}A^n\in\mathcal{L}(X).
\end{align*}
Now we verify that $S$ is genuinely the inverse of $I-A$, not just a formal expression. For each $N\in\mathbb N\cup\{0\}$, define
\begin{align*}
S_N=\sum_{n=0}^{N}A^n.
\end{align*}
Multiplying the finite sum gives the telescoping identity
\begin{align*}
(I-A)S_N=I-A^{N+1}.
\end{align*}
The same computation on the other side gives
\begin{align*}
S_N(I-A)=I-A^{N+1}.
\end{align*}
Since
\begin{align*}
\|A^{N+1}\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^{N+1}\to 0,
\end{align*}
we may pass to the limit in operator norm as $N\to\infty$. This yields
\begin{align*}
(I-A)S=I
\end{align*}
and
\begin{align*}
S(I-A)=I.
\end{align*}
Therefore $I-A$ has the two-sided inverse $S$, so $I-A$ is invertible in $\mathcal{L}(X)$.
[/guided]
[/step]
[step:Conclude that $T-\lambda I$ is invertible]
The factorization from the first step gives
\begin{align*}
T-\lambda I=(T-\lambda_0 I)(I-A).
\end{align*}
Both factors on the right are invertible bounded operators on $X$: the first by $\lambda_0\in\rho(T)$, and the second by the Neumann series argument above. Hence their product is invertible, with inverse
\begin{align*}
(T-\lambda I)^{-1}=(I-A)^{-1}R(\lambda_0,T).
\end{align*}
Therefore $T-\lambda I$ is invertible, so $\lambda\in\rho(T)$.
Since for every $\lambda_0\in\rho(T)$ the open disc
\begin{align*}
\{\lambda\in\mathbb C:|\lambda-\lambda_0|<\|R(\lambda_0,T)\|_{\mathcal{L}(X)}^{-1}\}
\end{align*}
is contained in $\rho(T)$, the resolvent set $\rho(T)$ is open in $\mathbb C$.
[/step]
