[proofplan]
Set $A:=T-\lambda I$ and $B:=T-\mu I$. Since $\lambda,\mu\in\rho(T)$, both $A$ and $B$ are invertible bounded operators on $X$, and their inverses are precisely the two resolvents. The algebraic identity $B-A=(\lambda-\mu)I$ gives the resolvent difference after multiplying by $A^{-1}$ on the left and $B^{-1}$ on the right. Multiplying in the opposite order gives the same difference with $B^{-1}A^{-1}$, and comparison yields commutativity.
[/proofplan]
[step:Introduce the two shifted operators and compute their difference]
Let $I:X\to X$ denote the identity operator. Define
\begin{align*}
A := T-\lambda I \in \mathcal{L}(X)
\end{align*}
and
\begin{align*}
B := T-\mu I \in \mathcal{L}(X).
\end{align*}
Because $\lambda,\mu\in\rho(T)$ and $\rho(T)$ is defined using the convention $T-\alpha I$, both $A$ and $B$ are invertible in $\mathcal{L}(X)$. Hence
\begin{align*}
A^{-1}=R(\lambda,T)
\end{align*}
and
\begin{align*}
B^{-1}=R(\mu,T).
\end{align*}
By subtracting the two definitions,
\begin{align*}
B-A=(T-\mu I)-(T-\lambda I)=(\lambda-\mu)I.
\end{align*}
[guided]
We first isolate the purely algebraic part of the argument. Let $I:X\to X$ be the identity operator, and define two bounded linear operators
\begin{align*}
A := T-\lambda I \in \mathcal{L}(X)
\end{align*}
and
\begin{align*}
B := T-\mu I \in \mathcal{L}(X).
\end{align*}
The hypothesis $\lambda,\mu\in\rho(T)$ means exactly that $T-\lambda I$ and $T-\mu I$ are invertible bounded operators on $X$, using the stated convention for the resolvent set. Therefore $A$ and $B$ are invertible in $\mathcal{L}(X)$, and the definition of the resolvent operator gives
\begin{align*}
A^{-1}=R(\lambda,T)
\end{align*}
and
\begin{align*}
B^{-1}=R(\mu,T).
\end{align*}
The point of introducing $A$ and $B$ is that their difference is a scalar multiple of the identity. Indeed, subtracting the two shifted operators gives
\begin{align*}
B-A=(T-\mu I)-(T-\lambda I)=(\lambda-\mu)I.
\end{align*}
This identity contains the factor $\lambda-\mu$ that appears in the desired formula.
[/guided]
[/step]
[step:Multiply the difference identity to obtain the resolvent formula]
Starting from
\begin{align*}
B-A=(\lambda-\mu)I,
\end{align*}
multiply on the left by $A^{-1}$ and on the right by $B^{-1}$. Associativity of composition in $\mathcal{L}(X)$ gives
\begin{align*}
A^{-1}(B-A)B^{-1}=(\lambda-\mu)A^{-1}IB^{-1}.
\end{align*}
The left-hand side simplifies as
\begin{align*}
A^{-1}(B-A)B^{-1}=A^{-1}BB^{-1}-A^{-1}AB^{-1}=A^{-1}-B^{-1}.
\end{align*}
The right-hand side simplifies as
\begin{align*}
(\lambda-\mu)A^{-1}IB^{-1}=(\lambda-\mu)A^{-1}B^{-1}.
\end{align*}
Therefore
\begin{align*}
A^{-1}-B^{-1}=(\lambda-\mu)A^{-1}B^{-1}.
\end{align*}
Substituting $A^{-1}=R(\lambda,T)$ and $B^{-1}=R(\mu,T)$ gives
\begin{align*}
R(\lambda,T)-R(\mu,T)=(\lambda-\mu)R(\lambda,T)R(\mu,T).
\end{align*}
[/step]
[step:Reverse the multiplication order to prove commutativity]
Again using
\begin{align*}
B-A=(\lambda-\mu)I,
\end{align*}
multiply on the left by $B^{-1}$ and on the right by $A^{-1}$. Then
\begin{align*}
B^{-1}(B-A)A^{-1}=(\lambda-\mu)B^{-1}IA^{-1}.
\end{align*}
The left-hand side simplifies as
\begin{align*}
B^{-1}(B-A)A^{-1}=B^{-1}BA^{-1}-B^{-1}AA^{-1}=A^{-1}-B^{-1}.
\end{align*}
The right-hand side simplifies as
\begin{align*}
(\lambda-\mu)B^{-1}IA^{-1}=(\lambda-\mu)B^{-1}A^{-1}.
\end{align*}
Thus
\begin{align*}
A^{-1}-B^{-1}=(\lambda-\mu)B^{-1}A^{-1}.
\end{align*}
Together with the identity proved in the previous step,
\begin{align*}
(\lambda-\mu)A^{-1}B^{-1}=(\lambda-\mu)B^{-1}A^{-1}.
\end{align*}
If $\lambda\ne\mu$, cancellation of the nonzero scalar $\lambda-\mu$ gives
\begin{align*}
A^{-1}B^{-1}=B^{-1}A^{-1}.
\end{align*}
If $\lambda=\mu$, this equality is immediate. Substituting $A^{-1}=R(\lambda,T)$ and $B^{-1}=R(\mu,T)$ yields
\begin{align*}
R(\lambda,T)R(\mu,T)=R(\mu,T)R(\lambda,T).
\end{align*}
This proves both the resolvent identity and the commutativity of the two resolvents.
[/step]
