[proofplan]
We prove the contrapositive form of the spectral containment. If $|\lambda|>\|T\|_{\mathcal L(X)}$, then the operator $\lambda I_X-T$ factors as $\lambda(I_X-\lambda^{-1}T)$, and the second factor is invertible by the Neumann series. This places every such $\lambda$ in the resolvent set of $T$, so no spectral value can lie outside the closed disk of radius $\|T\|_{\mathcal L(X)}$. The spectral radius bound then follows immediately from the definition of $r(T)$ as the supremum of $|\lambda|$ over $\lambda\in\sigma(T)$.
[/proofplan]
[step:Invert $I_X-\lambda^{-1}T$ by the Neumann series]
Let $\lambda\in\mathbb C$ satisfy $|\lambda|>\|T\|_{\mathcal L(X)}$. In particular $\lambda\ne 0$. Let $I_X:X\to X$ denote the identity operator on $X$, and define
\begin{align*}
S:X&\to X
\end{align*}
\begin{align*}
x&\mapsto \lambda^{-1}Tx .
\end{align*}
Then $S\in\mathcal L(X)$, and by homogeneity of the [operator norm](/page/Operator%20Norm),
\begin{align*}
\|S\|_{\mathcal L(X)}=|\lambda|^{-1}\|T\|_{\mathcal L(X)}<1.
\end{align*}
For each $N\in\mathbb N$, define the partial Neumann sum
\begin{align*}
R_N:X&\to X
\end{align*}
\begin{align*}
x&\mapsto \sum_{n=0}^{N} S^n x .
\end{align*}
Here $S^0:=I_X$, and $S^n$ denotes the $n$-fold composition of $S$ with itself for $n\ge 1$. For integers $M>N\ge 0$, [submultiplicativity of the operator norm](/theorems/1054) gives
\begin{align*}
\|R_M-R_N\|_{\mathcal L(X)}
\le \sum_{n=N+1}^{M}\|S^n\|_{\mathcal L(X)}
\le \sum_{n=N+1}^{M}\|S\|_{\mathcal L(X)}^n .
\end{align*}
Since $\|S\|_{\mathcal L(X)}<1$, the scalar geometric series converges, so $(R_N)_{N\ge 0}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $\mathcal L(X)$. Because $X$ is Banach, $\mathcal L(X)$ is Banach under the operator norm. Hence there exists $R\in\mathcal L(X)$ such that $R_N\to R$ in $\mathcal L(X)$.
For every $N\ge 0$, the finite telescoping identities give
\begin{align*}
(I_X-S)R_N=I_X-S^{N+1}
\end{align*}
and
\begin{align*}
R_N(I_X-S)=I_X-S^{N+1}.
\end{align*}
Also,
\begin{align*}
\|S^{N+1}\|_{\mathcal L(X)}\le \|S\|_{\mathcal L(X)}^{N+1}\to 0 .
\end{align*}
Passing to the limit in operator norm yields
\begin{align*}
(I_X-S)R=I_X
\end{align*}
and
\begin{align*}
R(I_X-S)=I_X.
\end{align*}
Thus $I_X-S$ is invertible in $\mathcal L(X)$, with inverse $R$.
[guided]
The goal is to show that the factor $I_X-\lambda^{-1}T$ is invertible. The condition $|\lambda|>\|T\|_{\mathcal L(X)}$ is designed exactly for this: after scaling $T$ by $\lambda^{-1}$, the resulting operator has norm strictly less than $1$.
Let $I_X:X\to X$ be the identity operator, and define the [bounded linear operator](/page/Bounded%20Linear%20Operator)
\begin{align*}
S:X&\to X
\end{align*}
\begin{align*}
x&\mapsto \lambda^{-1}Tx .
\end{align*}
Since $\lambda\ne 0$, this definition is valid. Since $T\in\mathcal L(X)$ and scalar multiplication preserves bounded linearity, $S\in\mathcal L(X)$. Its norm is
\begin{align*}
\|S\|_{\mathcal L(X)}=|\lambda|^{-1}\|T\|_{\mathcal L(X)}<1.
\end{align*}
We now build the inverse of $I_X-S$ explicitly. For each $N\in\mathbb N$, define
\begin{align*}
R_N:X&\to X
\end{align*}
\begin{align*}
x&\mapsto \sum_{n=0}^{N}S^n x ,
\end{align*}
where $S^0:=I_X$ and $S^n$ is the $n$-fold composition of $S$ for $n\ge 1$. These are finite sums of bounded linear operators, so $R_N\in\mathcal L(X)$.
To prove that the infinite series converges in operator norm, let $M>N\ge 0$. Using the triangle inequality in $\mathcal L(X)$ and submultiplicativity of the operator norm, we get
\begin{align*}
\|R_M-R_N\|_{\mathcal L(X)}
\le \sum_{n=N+1}^{M}\|S^n\|_{\mathcal L(X)}
\le \sum_{n=N+1}^{M}\|S\|_{\mathcal L(X)}^n .
\end{align*}
The right-hand side is the tail of a convergent geometric series because $\|S\|_{\mathcal L(X)}<1$. Therefore $(R_N)_{N\ge 0}$ is Cauchy in $\mathcal L(X)$. Since $X$ is Banach, the operator space $\mathcal L(X)$ is complete for the operator norm, so there exists $R\in\mathcal L(X)$ such that $R_N\to R$ in $\mathcal L(X)$.
It remains to verify that this limit is actually the inverse. The finite sums telescope:
\begin{align*}
(I_X-S)R_N=I_X-S^{N+1}.
\end{align*}
The same multiplication on the other side gives
\begin{align*}
R_N(I_X-S)=I_X-S^{N+1}.
\end{align*}
The error term tends to zero in operator norm because
\begin{align*}
\|S^{N+1}\|_{\mathcal L(X)}\le \|S\|_{\mathcal L(X)}^{N+1}\to 0.
\end{align*}
Passing to the operator-norm limit in the two telescoping identities gives
\begin{align*}
(I_X-S)R=I_X
\end{align*}
and
\begin{align*}
R(I_X-S)=I_X.
\end{align*}
Thus $R$ is both a right inverse and a left inverse for $I_X-S$, so $I_X-S$ is invertible in $\mathcal L(X)$.
[/guided]
[/step]
[step:Factor $\lambda I_X-T$ through the invertible Neumann factor]
Using the definition of $S$, we have
\begin{align*}
\lambda I_X-T=\lambda(I_X-S).
\end{align*}
Since multiplication by the nonzero scalar $\lambda$ is invertible, and $I_X-S$ is invertible in $\mathcal L(X)$ by the previous step, the product $\lambda(I_X-S)$ is invertible in $\mathcal L(X)$. An explicit inverse is
\begin{align*}
(\lambda I_X-T)^{-1}=(I_X-S)^{-1}\lambda^{-1}I_X .
\end{align*}
Therefore $\lambda I_X-T$ is invertible.
[/step]
[step:Exclude all points outside the operator-norm disk from the spectrum]
By definition, $\lambda\in\sigma(T)$ exactly when $\lambda I_X-T$ is not invertible in $\mathcal L(X)$. The previous step proves that every $\lambda\in\mathbb C$ satisfying $|\lambda|>\|T\|_{\mathcal L(X)}$ has $\lambda I_X-T$ invertible. Hence every such $\lambda$ belongs to the resolvent set of $T$, and therefore
\begin{align*}
\sigma(T)\subset \{\lambda\in\mathbb C:|\lambda|\le \|T\|_{\mathcal L(X)}\}.
\end{align*}
[/step]
[step:Take the supremum over spectral values to obtain the spectral radius bound]
By definition of the spectral radius,
\begin{align*}
r(T)=\sup\{|\lambda|:\lambda\in\sigma(T)\}.
\end{align*}
The containment just proved implies that every number in the set $\{|\lambda|:\lambda\in\sigma(T)\}$ is bounded above by $\|T\|_{\mathcal L(X)}$. Taking the supremum gives
\begin{align*}
r(T)\le \|T\|_{\mathcal L(X)}.
\end{align*}
This proves both assertions.
[/step]
