[proofplan]
The proof is componentwise. For each diagonal entry $\lambda_j$, positivity of the step size implies that $k\lambda_j$ remains in the closed left half-plane. A-stability therefore bounds the corresponding scalar amplification factor by one. The assumed diagonal decoupling gives a scalar recurrence on each component, and induction turns that recurrence into an explicit formula whose absolute value is non-increasing in $n$.
[/proofplan]
[step:Place each scaled diagonal eigenvalue in the A-stable region]
Fix an index $j\in\{1,\dots,m\}$. Define the scaled scalar
\begin{align*}
z_j:=k\lambda_j\in\mathbb{C}.
\end{align*}
Since $k>0$ is real and $\lambda_j\in\mathbb{C}_{-}$, we have
\begin{align*}
\operatorname{Re}(z_j)=\operatorname{Re}(k\lambda_j)=k\operatorname{Re}(\lambda_j)\le 0.
\end{align*}
Thus $z_j\in\mathbb{C}_{-}$. Since the method is A-stable, $\mathbb{C}_{-}\subset U$, so $z_j\in U$, and
\begin{align*}
|R(z_j)|\le 1.
\end{align*}
Equivalently,
\begin{align*}
|R(k\lambda_j)|\le 1.
\end{align*}
[/step]
[step:Use diagonal decoupling to obtain the scalar recurrence]
By the assumed componentwise decoupling property for diagonal linear systems, applying the method to $y'(t)=Ay(t)$ with step size $k$ gives, for every $n\in\mathbb{N}\cup\{0\}$ and every $1\le j\le m$,
\begin{align*}
Y_{n+1,j}=R(k\lambda_j)Y_{n,j}.
\end{align*}
This is exactly the asserted componentwise update formula.
[/step]
[step:Iterate the scalar recurrence and take absolute values]
Fix $j\in\{1,\dots,m\}$, and define the scalar amplification factor
\begin{align*}
a_j:=R(k\lambda_j)\in\mathbb{C}.
\end{align*}
The recurrence from the previous step is
\begin{align*}
Y_{n+1,j}=a_jY_{n,j}.
\end{align*}
We prove by induction on $n\in\mathbb{N}\cup\{0\}$ that
\begin{align*}
Y_{n,j}=a_j^nY_{0,j}.
\end{align*}
For $n=0$, this is $Y_{0,j}=a_j^0Y_{0,j}$, since $a_j^0=1$. If the identity holds for some $n\in\mathbb{N}\cup\{0\}$, then the recurrence gives
\begin{align*}
Y_{n+1,j}=a_jY_{n,j}=a_j(a_j^nY_{0,j})=a_j^{n+1}Y_{0,j}.
\end{align*}
Thus the formula holds for every $n\in\mathbb{N}\cup\{0\}$.
Taking absolute values and using multiplicativity of the complex modulus gives
\begin{align*}
|Y_{n,j}|=|a_j|^n|Y_{0,j}|.
\end{align*}
From the A-stability estimate in the first step, $|a_j|=|R(k\lambda_j)|\le 1$. Hence $|a_j|^n\le 1$ for every $n\in\mathbb{N}\cup\{0\}$, and therefore
\begin{align*}
|Y_{n,j}|\le |Y_{0,j}|.
\end{align*}
Since $j$ was arbitrary, the decay estimate holds for every component.
[guided]
Fix a component index $j\in\{1,\dots,m\}$. The point of working componentwise is that the diagonal decoupling assumption has reduced the vector method to a scalar recurrence. Define the scalar amplification factor
\begin{align*}
a_j:=R(k\lambda_j)\in\mathbb{C}.
\end{align*}
For this fixed component, the recurrence is
\begin{align*}
Y_{n+1,j}=a_jY_{n,j}
\end{align*}
for every $n\in\mathbb{N}\cup\{0\}$.
We first solve this recurrence exactly. We claim that
\begin{align*}
Y_{n,j}=a_j^nY_{0,j}
\end{align*}
for every $n\in\mathbb{N}\cup\{0\}$. The base case is $n=0$: by the convention $a_j^0=1$, the identity reads
\begin{align*}
Y_{0,j}=a_j^0Y_{0,j}.
\end{align*}
Now assume the identity holds at some time index $n\in\mathbb{N}\cup\{0\}$. Substituting this inductive hypothesis into the recurrence gives
\begin{align*}
Y_{n+1,j}=a_jY_{n,j}=a_j(a_j^nY_{0,j})=a_j^{n+1}Y_{0,j}.
\end{align*}
This proves the explicit formula by induction.
Now we convert the explicit formula into a decay estimate. Taking absolute values and using the multiplicative property $|uv|=|u||v|$ for complex numbers gives
\begin{align*}
|Y_{n,j}|=|a_j^nY_{0,j}|=|a_j|^n|Y_{0,j}|.
\end{align*}
Because $k>0$ is real and $\lambda_j\in\mathbb{C}_{-}$, the scaled value $k\lambda_j$ satisfies
\begin{align*}
\operatorname{Re}(k\lambda_j)=k\operatorname{Re}(\lambda_j)\le 0.
\end{align*}
Hence $k\lambda_j\in\mathbb{C}_{-}$. A-stability says that $\mathbb{C}_{-}\subset U$ and $|R(z)|\le 1$ for every $z\in\mathbb{C}_{-}$, so $R(k\lambda_j)$ is defined and
\begin{align*}
|a_j|=|R(k\lambda_j)|\le 1.
\end{align*}
Therefore $|a_j|^n\le 1$ for every $n\in\mathbb{N}\cup\{0\}$. Substituting this into the previous identity yields
\begin{align*}
|Y_{n,j}|\le |Y_{0,j}|.
\end{align*}
Because the index $j$ was arbitrary, the same argument applies to every diagonal component.
[/guided]
[/step]
