$T\{x[n]\}=\sum _{k=\text{min}(n,{n}_{0})}^{\text{max}(n,{n}_{0})}x[k]$

for some integer constant ${n}_{0}$.

Intuitively, it's unstable, and it can be easily proven by a counterexample if $x[n]$ is the unit step and ${n}_{0}=0$.However, the TA tried to use a general proof by invoking the triangle inequality. Assuming $x[n]$ is bounded (or $|x[n]|\le M<\mathrm{\infty}$), he said:

$\left|T\{x[n]\}\right|=\left|\sum _{k=\text{min}(n,{n}_{0})}^{\text{max}(n,{n}_{0})}x[k]\right|\le \sum _{k=\text{min}(n,{n}_{0})}^{\text{max}(n,{n}_{0})}|x[k]|\le (|n-{n}_{0}|+1)M$

Obviously the right-hand side is unbounded, as it goes to infinity with increasing $n$, but to me it doesn't seem to imply that the system on the left-hand side is unbounded (because of the inequality).

My question is, can his attempt be augmented to show that the left-hand side is also unbounded? Or is a counter-example the only way to prove it?

for some integer constant ${n}_{0}$.

Intuitively, it's unstable, and it can be easily proven by a counterexample if $x[n]$ is the unit step and ${n}_{0}=0$.However, the TA tried to use a general proof by invoking the triangle inequality. Assuming $x[n]$ is bounded (or $|x[n]|\le M<\mathrm{\infty}$), he said:

$\left|T\{x[n]\}\right|=\left|\sum _{k=\text{min}(n,{n}_{0})}^{\text{max}(n,{n}_{0})}x[k]\right|\le \sum _{k=\text{min}(n,{n}_{0})}^{\text{max}(n,{n}_{0})}|x[k]|\le (|n-{n}_{0}|+1)M$

Obviously the right-hand side is unbounded, as it goes to infinity with increasing $n$, but to me it doesn't seem to imply that the system on the left-hand side is unbounded (because of the inequality).

My question is, can his attempt be augmented to show that the left-hand side is also unbounded? Or is a counter-example the only way to prove it?